A Countable Powerset -- Where am I wrong? Cantor's Theorem seems pretty airtight to me. So what is wrong with the following reasoning?
Consider $X = \mathcal{P}(\mathbb{N})$. We say $X$ contains uncountably many sets. But the only sets that exist are those asserted to exist by the axioms. So if we consider each of the axioms that assert more sets, starting with $\emptyset$, we have one set. The pairing axiom can only produce countably many more. Union gives us up to one more set per set in existence. Powerset also gives us up to one more set per set in existence. Separation and replacement gives us countably many more sets per set in existence. Infinity gives us more more set. Foundation gives us one more set per set in existence. Choice also only gives us one more set per set in existence.
It seems to me that the axioms only ever define countably many sets. So when we consider all the subsets of a set, there might be uncountably many things that satisfy the condition of being a subset, but only countably many of them are possibly sets. If sets can only contain sets, $\mathcal{P}(\mathbb{N})$ must be countable.

I'm still not getting the argument about parameters coming from somewhere. I understand the notion of parameters in the context of definability in a structure, but this question is about a theory (ZFC), whose axioms are well formed formulas in the language of set theory ($\forall,\land,\neg,\in$). Thus technically all the separation and replacement axioms do not contain any parameters, right?
Considering the example set:
$$ X = \{ (n,k) | 2^{\aleph_n} = \aleph_k \} $$
When all the subsets of $\mathbb{N}^2$ are added, surely $X$ is among them? We can say with certainty is that $X$ exists at this stage (because it is definable...or perhaps we can only say there is a plethora of possibilities for $X$ in existance?), but we would be unable to enumerate the pairs (ie say which subset of $\mathbb{N}^2$ it is) without making more assertions about the continuum.
I like this idea that there can exist subsets of $\mathbb{N}$ that are undefinable without parameters that come from sets of a higher rank. Perhaps there is a forcing notion that makes $X$ one of these in a particular model. But are there uncountably many of them?
 A: You hit some sort of circularity with separation.
While there are only countably many formulas (and that countable collection is not really countable from the standpoint of set theory. It's nonexistent, since it's a part of the meta-theory); these formulas are allowed to have parameters.
All you really show here is that a lot of the subsets of $\Bbb N$ will be definable using separation axioms of the form $\varphi(x,p):= x\in p$ where $p$ is a parameter, and $A=\{x\in\Bbb N\mid\varphi(x,A)\}$. It's a bit of circular, but not if you really think about it.
But there is another issue here. Separation axioms don't rely on "previously added sets" in the hierarchal order of things. Note that $\{(n,k)\mid 2^{\aleph_n}=\aleph_k\}$ is a definable subset of $\Bbb N^2$ (and so by encoding it defines a subset of $\Bbb N$), but it's not really added after adding $\Bbb N$. It will only be added after we've added a lot more, namely all the $\aleph_n$'s and all their power sets.
And we can add all sort of crazy definitions like that. The freedom to use parameters, and to use formulas which address the entire set theoretical universe and not just $\Bbb N$, show that there are uncountably many subsets. The power set axiom just ensures that there is just "set many" of them.
What's even more important is to understand that the axioms don't "define" the structure". Instead the structure is given and we check that it satisfies the axioms. Does your argument mean that any model of a countable language is countable, just because? Not really. It just shows that a countable language can define only countably many elements in a model (unless you allow parameters).
Let me point a similarity with the von Neumann construction of $V$ (or really, any other hierarchal construction). The sets come from somewhere. They already exist. We just show that we can write the universe as a limit of a hierarchy of sets. Similarly in your question, the sets already exist, the axioms don't bring them to life, they just allow us to show that particular sets exist. 
A: I believe the crux of the issue is grasping that Cantor's theorem requires the bijection to exists in the model.
When we say $X = \mathcal{P}(\mathbb{N})$ is "uncountable", it simply means that there does not exist a bijection from $\mathbb{N}$ to $X$ in whatever model. So, if we let $M$ be a countable model of ZFC, the cardinality of $\mathcal{P}(\mathbb{N})$ is"countable" in a metalanguage sense of the term -- each of the subsets are in a one to one correspondence with the "countable" $\varphi$ that define them. However, there exists an injective function $f_i$ from $\mathbb{N}$ to $X$ such that $f_i \in M$, but there does not exist a bijective function $f_b$ from $\mathbb{N}$ to $X$ such that $f_b \in M$ (else $M$ would not be a model of ZFC).
I see now that my question was, in fact, equivalent to the Skolem Paradox. In my question I assert that the only sets that exist are those that provably exist from the axioms, but this assertion is not provable from ZFC. However, it is true in a model constructed by the downward Lowenheim-Skolem theorem. Inside such a model, the powerset of the natural numbers is uncountable, even though it is countable in whatever meta-model the construction of this model took place.
