Simple example of a moduli space? This relates to my previous question: Is $\zeta_{\Bbb R^n}$ a space of moduli spaces?
Moduli Space
I asked two professors about this but they said they didn't know enough about moduli spaces to help.
I just want to look at a simple example, $\zeta_{\Bbb R^2}.$ The $x-$coordinates in the space are the solutions to polynomials of the form: $x^T-(1-x)^S=0.$ Plugging the solution back into the original function (given in the previous post) gives the $y-$ coordinate in the space. So basically each point encodes the solutions of two algebro-geometric problems.
Maybe I'm missing something, but wouldn't this be a simple example of a moduli space?
 A: $\newcommand{\E}{\mathcal{E}^\mathrm{univ}}$$\newcommand{\mf}[1]{\mathfrak{#1}}$This was originally a comment, but it grew too long. It's pretty stream of conscience though, so let me know if something needs clarification.
From this post and the previous it really seems as though you are struggling to provide great meaning to the notion of moduli space. It took me a long time to realize this, but it's not exactly worth chasing a rigorous deifnition early on in your mathematical pursuits. In fact, everything is a moduli space: it's a moduli space for its own points (see a funny anecdote of Kisin on this topic at 59:50 of this). 
To actually be able to say "This is a moduli space" or "This is not a moduli space" requires more sophistication than you are assuming. In particular, the first pass definition of a moduli space is "$X$ is a moduli space for some set of objects $S$ if there is a natural bijection between $X$ and $S$". Of course, this is wholly unsatisfying since 'natural' is a super vague weasel word here. In particular, there is nothing that uniquely characterizes $X$ by this definition (lots of spaces could have 'natural bijections' with the elements of $S$).
It turns out that, following the Grothendieckian/Yonedian philosophy, the way that one can make the above definition precise is to require that the objects of $S$ not only make sense in an absolute sense but also 'in families' over base objects $T$. Then, $X$ becomes a moduli space for objects $S$ if not only are the points of $X$ in bijection with $S$ but the points of $X\times T$ are in bijection with the families of objects of $S$ living over $T$. More rigorously, a moduli space for a (contravariant) functor $\mathcal{F}$ on a category $\mathcal{C}$ is an object $X$ of $\mathcal{C}$ such that $\mathcal{F}$ is isomorphic to the functor $\mathrm{Hom}(-,X)$. 
So, starting with a space and asking if it's a moduli space is somewhat contentless since yes, it's a moduli space for the functor $\mathrm{Hom}(-,X)$. What you need to do instead is start with a moduli problem, a way of associating to a base object $T$ some notion of families of $S$-objects over $T$, and then ask whether this association (this functor) is representable. 
Something that is obscured in the above is that whether $X$ is a moduli space or not really heavily depends not only on the moduli problem $\mathcal{F}$ but on the category $\mathcal{C}$. In particular, if you have some moduli problem $\mathcal{F}$ on a category $\mathcal{C}$ you can restrict it to a subcategory $\mathcal{C}'$ of $\mathcal{C}$. Then, it's possible that $\mathcal{F}$ is not representable on $\mathcal{C}$ but is on $\mathcal{C}'$ and vice versa. 
So, let's come back to your question. To really ask whether your space $\zeta_{\mathbb{R}^2}$ is a moduli space in a rigorous sense you need to specify what category you are interested in. I would guess from context you mean that $\zeta_{\mathbb{R}^2}$ should be a moduli space in the category of smooth manifolds. Next, we need to write down what the moduli problem should be. It seems as though you want it to be the solutions to some polynomial equation $p(x,y)$. The problem is that to really make this a moduli problem we need to have an interpretation for these solutions for any manifold $M$. The problem a priori is that I don't know how to interpret "the solutions to $p(x,y)=0$ on $M$" when $p(x,y)$ is not a sensical function on $M^2$. So, it doesn't really seem obvious to me as to how to put your space into a rigorous framework. So, while your space $\zeta_{\mathbb{R}^2}$ may be a 'moduli space' in the sense that its points are naturally in bijection with some set of solutions of some polynomial equation, this is (as discussed above) fairly contentless since since natural bijections don't really characterize the space at all.
We can adapt ideas of what you're saying to get a true moduli space. Namely, let's think of replacing manifolds by rings. Let's fix a polynomial $p(x,y)$ with integer coefficients. We have a moduli problem on the category $\mathsf{Ring}$ of all rings that associates to a ring $R$ the set $\{(x,y)\in R^2:p(x,y)=0\}$. This is, in fact, representable by a moduli space. Really if we want things to work out we should be working in the category of affine schemes $\mathsf{Aff}$ (the opposite category of $\mathsf{Ring}$) which (since you've tagged this as algebraic geometry) I assume you're OK with. Then our moduli problem associates to $\mathrm{Spec}(R)$ the set $\{(x,y)\in R^2:p(x,y)=0\}$. The representing space for this moduli problem is $\mathrm{Spec}(\mathbb{Z}[x,y]/(p(x,y))$ since there is a (functorial) bijection between $\mathrm{Hom}(\mathrm{Spec}(R),\mathrm{Spec}(\mathbb{Z}[x,y]/(p(x,y)))$ and $\{(x,y)\in R^2:p(x,y)=0\}$ as I leave you to check. 
In some sense, this is almost tautological, and so you may want to consider a moduli problem that less obviously representable or, what is even more instructive, a moduli problem that is not representable. The classical example is the moduli problem of elliptic curves. Namely, to a ring $R$ one can associate the set of isomorphism classes of elliptic curves over $R$ (whose definition is a little involved, but you can just think of as a family of elliptic curves indexed by hte points of $\mathrm{Spec}(R)$). This then is a moduli problem that is not representable (the proof of which is very slightly complicated, although fairly intuitive, so I won't recount it here). There is no affine scheme $X$, or even scheme, for which the isomorphism classes of elliptic curves over $R$ are in (functorial) bijection with $\mathrm{Hom}(-,X)$. This highlights the confusing fact that people often times call $\mathbb{A}^1_\mathbb{C}$ the 'moduli space of elliptic curves over $\mathbb{C}$' because the elliptic curves over $\mathbb{C}$ are in 'natural bijection' (there's that weasel word again) with the isomorphism classes of ellitpic curves over $\mathbb{C}$. 
As a nice example of a closely related moduli problem that does have a solution, you can consider the category $\mathcal{C}$ of complex analytic spaces (or even complex manifolds if you like) and the functor which associates to such a space $T$ the isomorphism classes of elliptic curves $E$ over $T$ together with a trivialization $\psi$ (i.e. isomorphism with $\mathbb{Z}^2$) of their homology $H_1(E)$ (in families one has to interpret this in terms of higher pushforward sheaves). It turns out that this moduli problem is representable by the upper and lower half-planes $\mathfrak{h}^{\pm}=\mathbb{C}-\mathbb{R}$. Namely, there is a functorial bijection between elliptic curves with homology trivialization over a space $T$ and the maps $\mathrm{Hom}(T,\mathfrak{h}^{\pm}$) (e.g. see Section 6 of this). This is highly contentful since it's not at all obvious that such a space (especially one as simple as $\mathfrak{h}^{\pm}$) should dictate the structure of such a complicated moduli problem. 
Hopefully this clarifies things a bit.
EDIT: There was a question asked in the comments that I think is worth on expanding here. Namely, while I did caution you against worrying too much about moduli spaces early in your mathematical career (as I did) I would also say that I was brought comfort knowing there was rigorous definitions that I could eventually learn--that moduli spaces weren't some ethereal object one had to commune with using Ouija boards. So, it may be helpful to give a rigorous answer to the question.
The question is, if I understood it correctly, the following. Let $\mf{h}^+$ denote the upper half-plane. Note then that one has a 'natural map' from $\mf{h}^+$ to the set $S_0$ of elliptic curves over $\mathbb{C}$ up to isomorphism. Indeed, to an element $\tau\in\mf{h}^+$ one obtains the lattice $\Lambda_\tau:=\mathbb{Z}+\mathbb{Z}\tau$ and then one obtains the elliptic curve $E_\tau:=\mathbb{C}/\Lambda_\tau$. Thus, the mapping $\mf{h}^+\to S_0$ I am alluding to is the map $\tau\mapsto E_\tau$. 
Moreover, one can show that $E_\tau\cong E_{\tau'}$ if and only if $\tau$ and $\tau'$ lie in the same orbit of the action of $\mathrm{SL}_2(\mathbb{Z})$ on $\mf{h}^+$ given by 
$$\begin{pmatrix}a &b\\ c &d\end{pmatrix}z:=\frac{az+b}{cz+d}$$
This then suggests that the induced map
$$\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})\to S_0:\tau\mapsto E_\tau$$
is well-defined. In fact, one can show it's a bijection. 
Thus, given the 'naturalness' of the above bijection, one is tempted to make the following claim:

(Incorrect) Claim: The analytic space $\mf{h}^+/\mathrm{SL}_2(\mathbb{C})$ is the moduli space of elliptic curves over $\mathbb{C}$.

But, as it says on the tin, this claim is incorrect. To see this, let's unravel what it means to say that the association 
$$T\mapsto \{\text{Elliptic curves over }T\}/\text{iso.}$$
is represented by $\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$. Once you get past all the language in Yoneda's lemma, what it really says is this. It says that there is some universal elliptic curve $\E\to \mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$. What do we mean by universal? We mean that for any complex analytic space $T$ we get a map
$$\mathrm{Hom}(T,\mf{h}^+/\mathrm{SL}_2(\mathbb{Z}))\to \{\text{Elliptic curves over }T\}/\text{iso.}$$
which takes a map $t:T\to \mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$ to the elliptic curve $t^\ast \E$ (the pullback of the universal elliptic curve). Being universal means that this map is always a bijection. This is the point of the Yoneda philosophy, and the point of moduli spaces--they carry a universal object.
So, why is this a problem for $\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$? Why can it not carry a universal elliptic curve? Simple! Because, in fact, every elliptic curve over $\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$ is constant. It looks like $E_0\times (\mf{h}^+/\mathrm{SL}_2(\mathbb{Z}))$. Evidently such a constant elliptic curve cannot be universal since, for example, if you take $T$ to be a point (i.e. $\mathrm{Spec}(\mathbb{C})$ or $\mathrm{Sp}(\mathbb{C})$ depending if you're working in the algebraic or complex analytic category) then we would need that the map $t\mapsto t^\ast(E_0\times (\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})))$ is a bijection
$$\mathrm{Hom}(\text{pt},\mf{h}^+/\mathrm{SL}_2(\mathbb{Z}))\to \{\text{Elliptic curves over }\mathbb{C}\}/\text{iso.}$$
But, evidently $t\mapsto t^\ast(E_0\times (\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})))$ is always just $E_0$! So, this is preposterous.
So, the only question remains is why every elliptic curve over $\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$ is constant. The beauty is, let us prove this with a moduli space argument. 
Indeed, suppose that $f:\mathcal{E}\to \mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$ is an elliptic curve. Note that while its obscured by its presentation $\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$ (when you endow it with a complex analytic structure in the usual way, which requires some care, as in Diamond and Shurman's text on modular forms) is just $\mathbb{A}^1(\mathbb{C})$. Note then that since $\mathbb{A}^1(\mathbb{C})$ is simply connected, the local system $(R^1f_\ast\underline{\mathbb{Z}})^\vee$ (the relative homology sheaf) must be trivial. So, choose an isomomorphism $\psi$ between this sheaf and $\underline{\mathbb{Z}}^2$. Note then that the pair $(\mathcal{E},\psi)$ is precisely an elliptic curve over $\mathbb{A}^1(\mathbb{C})$ with a trivialization of homology as in the my pre-Edit post. 
But, I told you that the moduli space of such objects is $\mf{h}^{\pm}$. So, by definition, we know that there is some universal pair $(\mathcal{C}^\mathrm{univ},\psi^\mathrm{univ})$ (of an elliptic curve with homology trivialization) and some map $t:\mathbb{A}^1(\mathbb{C})\to \mf{h}^{\pm}$ such that 
$$(\mathcal{E},\psi)\cong t^\ast(\mathcal{C}^\mathrm{univ},\psi^\mathrm{univ})$$
But, guess what? The only maps from $\mathbb{A}^1(\mathbb{C})\to \mf{h}^{\pm}$ are constant! Indeed, this follows Picard's Little Theorem! So, since $t$ is constant it's easy to see that $t^\ast\mathcal{C}^\mathrm{univ}$ is constant--it looks like $E_0\times\mathbb{A}^1(\mathbb{C})$ for some $E_0$ as desired.
So, in the above you saw a rigorous way to show that a space whose 'points are naturally in bijection with some set' is not a moduli space and, moreover, saw how to utilize moduli spaces in the proof!
NB: It's worth noting that while $\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$ is not a moduli space, or what is more precisely called a fine moduli space, it is something much weaker--a so-called coarse moduli space. Intuitively a functor/moduli problem $\mathcal{F}$ has $X$ as a coarse moduli space if there is a map $\mathcal{F}\to X$ which is a 'best approximation of $\mathcal{F}$ by a space' (i.e. $\mathcal{F}\to X$ is initial amongst maps from $\mathcal{F}$ to actual spaces) and which is a bijection on algebraically closed fields (or, in some sense, is a bijection on points). This is probably what people mean when they say that $\mf{h}^+/\mathrm{SL}_2(\mathbb{Z})$ is a moduli space (although I think many texts are just imprecise). This notion of coarse moduli space is, in some sense, even more sophisitcated than the notion of fine moduli space, so I'd ignore it for now if it confuses you.
