Basically, there is a theorem of first-order logic that says that no L-structure whose domain is infinite can be axiomatized up to isomorphism, so in particular there is no set $\Phi$ of formulas that attempts to axiomatize first-order arithmetic where all structure $\mathfrak{A}$ such that $\mathfrak{A}\vDash\Phi$ is isomorphic. Because of this, we know that there are non-isomorphic models that satisfy the Peano axioms. But by choosing a model $M\vDash\text{ZFC}$ we know that $M$ manages to prove that every (internal, i.e., in $M$) Peano structure that satisfies all $\Phi$ are isomorphic, which is an idiosyncrasy of each $\text{ZFC}$ model. So basically each $M$ thinks they have the canonical structure of arithmetic, whereas between them these are non-isomorphic, but this can only be seen "from the outside", therefore there are at least as many $\text{ZFC}$ models as non-isomorphic Peano arithmetic models.
So this is where my doubt arises, if the models of Peano arithmetic that I can build in any $\text{ZFC}$ model is always a subset of all Peano structures, then using ZFC as a metatheory to do first-order logic does not mean that I'll be doing a less expressive first-order logic? Or less powerful in some sense, since I'm never going to be able to build these non-standard models of arithmetic, or at least, the non-isomorphism between them?