this might be a silly question, but I was wondering: PA cannot prove its consistency by the incompleteness theorems, but we can "step outside" and exhibit a model of it, namely $\mathbb{N}$, so we know that PA is consistent.
Why can't we do this with ZFC? I have seen things like "if $\kappa$ is [some large cardinal] then $V_{\kappa}$ models ZFC", but these stem from an "if".
Is this a case of us not having been able to do this yet, or is there a good reason why it is simply not possible?
The problem is that, unlike the case for PA, essentially all accepted mathematical reasoning can be formalized in ZFC. Any proof of the consistency of ZFC must come from a system that is stronger (at least in some ways), so we must go outside ZFC-formalizable mathematics, which is most of mathmatics. This is just like how we go outside of PA-formalizable mathematics to prove the consistency of PA (say, working in ZFC), except that PA-formalizable mathematics is a much smaller and relatively uncontroversial subset of mathematics. Thus it is a common view that the consistency of PA is a settled fact whereas the consistency of ZFC is conjectural. (As I mentioned in a comment, as a gross oversimplification, “settled mathematical truth (amongst mainstream classical mathematicians)” roughly corresponds to “provable in ZFC”... Con PA is whereas Con ZFC is not.)
As for proving the consistency of ZFC, as you mention, we could go to a stronger system where there is an axiom giving the existence of inaccessible cardinals. Working in Morse-Kelley is another possibility. In any case, you are in the same position with the stronger system as you were with ZFC: you cannot use it to prove its own consistency, and since you’ve made stronger assumptions, you’re at a higher “risk” of inconsistency.