Prove that the upper half plane of Cartesian plane is not an affine variety The question is stated as follows,

Q. Let $R=\{(x,y):y>0\}$ be the upper half plane. Prove that $R$ is not an affine variety.

My attempt was simply first assuming that it is an affine variety and seek for a contradiction. However, I could't proceed further and the solution is as follows

A. If $R=V(f_1,...,f_s)$ then each $f_i$ vanishes on $Z^2_{M_i+1}$ (Note:Though the format is not $\mathbb{Z}$ in the solutions for this "Z" but it probably means "integers") where $M_i$ is the largest power of $x$ or $y$ in $f_i$.

My request is, could someone render the above statement in a somewhat more colloquial or a simpler way? What I do not understand is, why "$R=V(f_1,...,f_s)$"  implies "each $f_i$ vanishes on $Z^2_{M_i+1}$." If $M_i$ is the largest power of $x$ or $y$, why $M_i+1$?
Perhaps another slight confusion is, I am familiar with notation $\mathbb{Z}^n_{\geq 0}$ which means, for the subscript, "elements of the $n$-tuples are $\geq 0$" but here, $Z_{M_i+1}$ means...what? Does it mean all elements are $\geq M_i+1$ though it does not have the "$\geq$" sign?

(continued solution) Thus each $f_i$ is the zero polynomial and so $V=R^2$. Since $R^2 \neq R$ we have a contradiction.

I believe this statement has to do with $f_i$ having "more solutions than its degree $M_i$(implies $f_i$ must be the zero polynomial)" but, since I am not clear with the statements above, I am unsure. And finally, I don't see how this leads to the conclusion  "$V=R^2$."
Essentially, I am full of question marks here. I'll try the thinking on my own once I see what the solution is trying to say. Right now, I don't even see what it's trying to convey due to the lack of my knowledge.
So it would be great if someone can elaborate this solution based on my confusion stated above. Thank you, please comment if anything is unclear
 A: If you want to prove some set is not affine varieties, there would be two main steps:
First, find out what would it be if it vanishes in the set: In this process, you will need to use the property that a nonzero polynomial can only have finitely many distinct roots, while there are infinitely many roots in the upper half plane, the only possibility is that it is the zero polynomial. Therefore, it would be the whole plane of real numbers.
Second, find out the contradiction: there are so many counter examples in the other half plane, which should be in the plane but excluded. In short, we just say that R is not equal to R^2.
A: As per Exercise 6b. of Chapter 1, Section 1 (page 5) of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms, the notation $\mathbb{Z}^n_{M+1}$ denotes the set of all points of $\mathbb{Z}^n$, all of whose coordinates lie between $1$ and $M+1$. So, for instance, $(1,1,2)$ lives in $\mathbb{Z}^3_{1+1}$, but $(0,1,2)$ does not (so, $M=1$ and $n=3$ in this example). Cox, Little, and O'Shea ask the reader to prove that if $f$ vanishes at all points of $\mathbb{Z}_{M+1}^n$, then $f$ is the zero polynomial. The exercise you would like to solve asks you to apply this result. 
For each of the polynomials $f_i$ in the vanishing set $R=V(f_1,...,f_s)$, let $M_i$ denote the largest power of $x$ or $y$ appearing in $f_i$. Then, by the aforementioned exercise, since each $f_i$ vanishes on each point of the upper half plane, each $f_i$ vanishes at the integer lattice points of $\mathbb{Z}_{M_i+1}^2$, and so each $f_i$ is the zero polynomial. But this a contradiction, since we're assuming they only vanish on the upper half plane!
