Gödel's second incompleteness theorem tells us that ZFC (if it is consistent) cannot prove that ZFC is consistent. Therefore, ZFC (if it is consistent) cannot prove that there is a model of ZFC, either. Thus we do not have a proof that no model exists, but we do have a proof that trying to find such a model is a pointless pursuit.
If you make stronger assumptions, such as assuming the existence of large cardinals, then you can prove that models of ZFC exist, and you can study them.
This is a surprisingly subtle question!
Right off the bat, we have an obvious observation: if $\mathcal{M}$ is a "potential $\omega$" (= is isomorphic to the $\omega$ of some model of $\mathsf{ZFC}$), then $\mathcal{M}$ must satisfy all the arithmetic consequences of $\mathsf{ZFC}$. This set of sentences, call it "$\mathsf{ZFC_{arith}}$," is not very well understood but we can prove some basic things about it; see e.g. this old answer of mine.
That said, it's also worth noting that under a mild assumption we can whip up a formula that $\mathsf{ZFC}$ proves defines a model of $\mathsf{PA}$ which has no additional restrictions - see here - but this is quite different from the usual way arithmetic is implemented in $\mathsf{ZFC}$, which you are asking about.
But is that enough? That is, if $\mathcal{M}\models\mathsf{ZFC_{arith}}$, is $\mathcal{M}$ a potential $\omega$?
To see the issue, consider the following very artificial example. Let's say we replace the interesting arithmetic structure of the naturals with just the single equivalence relation "has the same parity as." Basically, this turns the naturals into two infinite blobs. Now consider a structure $\mathcal{A}$ in this language consisting of two infinite blobs of different cardinalities. Since the theory of two infinite blobs is complete, $\mathcal{A}$ satisfies everything $\mathsf{ZFC}$ proves about "the naturals as a pair of blobs." However, $\mathcal{A}$ is not isomorphic to any model of $\mathsf{ZFC}$'s "blobby naturals." This is because the $\mathsf{ZFC}$-theorem "the even and odd naturals have the same cardinality" does have consequences for the isomorphism type of "potential blobby naturals" but is not captured at the first-order level.
In fact, $\mathsf{ZFC_{arith}}$ falls well short of "potential $\omega$-ness" in a completely unfixable way. Barwise/Schlipf proved the following:
For a countable nonstandard $\mathcal{M}\models\mathsf{PA}$, the following are equivalent:
(Strictly speaking, they proved that any structure appearing in some non-$\omega$-model of $\mathsf{ZF}$ is recursively saturated. This gives the top-to-bottom direction, with the bottom-to-top direction being folklore if I understand the history correctly. Note that countability is actually only needed in the bottom-to-top direction!)
The omitting types theorem then prevents any first-order theory from capturing potential $\omega$-ness. (Thanks to Andreas Blass for pointing this out to me in the comments below!) Depending on whether you like recursive saturation, this may or may not answer your question in the specific case of countable models. Unfortunately (or interestingly), uncountable models pose much more difficulties, and I believe no characterization is currently known even of the size-$\aleph_1$ models of $\mathsf{PA}$ which are isomorphic to some $\mathsf{ZFC}$-models' version of the naturals. (But that shouldn't be too surprising, since even the ordertypes of size-$\aleph_1$ models of $\mathsf{PA}$ aren't classified yet.)
As a coda, I vaguely recall an old paper (by H. Friedman?) on the following problem: given a first-order theory $T$, when does a tuple of formulas $\Phi$ have the property that, if we let $T_\Phi$ be the set of sentences $T$ proves holds of the structure interpreted by $\Phi$, then for all $\mathcal{A}\models T_\Phi$ there is some $\mathcal{B}\models T$ with $\mathcal{A}\cong \Phi^\mathcal{B}$? Basically, these $\Phi$s should not be "missing" any structural data (e.g. the "blobby naturals" example above was missing the structural data of the $n\mapsto n+1$ bijection). I can't seem to find the paper at the moment, however. My recent mathoverflow questions 1, 2 address this general theme, current absence of a reference notwithstanding.
Best Answer
Yes, that is correct. The choice of your mathematical universe will affect the objects inside the universe.
If you believe that inaccessible cardinals are inconsistent, but I believe that they are fine, then we fundamentally disagree on $\Bbb N$.
But the standard model is called the standard model, because inside a universe of set theory there is exactly one standard model. And I think that no foundational theory is considered foundational if it does not have some sort of uniqueness theorems for $\Bbb N$ (and $\Bbb R$, while we're at it).
So yes, we can disagree on our meta-theory which in turn means we might disagree on $\Bbb N$. But we also agree that it has to be the unique models of $\sf PA$ which is well-ordered. And then we can ask, what are the things that we can both definitely prove from $\sf PA$, and in this grey area, there is a lot of interesting mathematics, and arguably where most of our efforts are concentrated (for better or worse).