Is there a binary relation that could possibly be a model of (e.g.) ZF? (Note: Though I am talking about $ZF$ here for concreteness, my question applies equally to $ZFC$ and presumably to other popular set theories as well, though I understand it may not have the same answer in all of them.)
If $ZF$ is consistent then it has a model, which means there exists some particular relation set $(M, \in)$ which is that model. Thanks to incompleteness we will not be able to prove the existence of this model within $ZF$ itself, but this does not necessarily mean the set itself is completely inaccessible. In principle we could imagine that $(M, \in)$ might be constructible or at least known to exist as a relation set within ZF, and it would be only the proof that it constitutes a model which eludes us.
We can state this imagined scenario as the following proposition, given within $ZF$:

There exists a relation set $(M, \in)$ such that it cannot be disproved that $M$ is a model of $ZF$.

Is the truth status of this proposition known?
 A: Here is a concrete example for you.

*

*If $\sf ZF$ is consistent, then $({\sf ZF})^L$ is consistent. Namely, the interpretation of the axioms in the inner model $L$ is consistent as well. Mainly because $L$ and $V$ have the same integers, therefore the same recursively constructed things, and therefore the same first-order logic, the same axioms of $\sf ZF$, and the same proofs.


*If $\sf ZF$ is consistent, there is an $L$-minimal $E\subseteq\omega\times\omega$ such that $(\omega,E)\models\sf ZF$.
Now simply let $E$ be empty if $\sf ZF$ is inconsistent and the $L$-minimal witness otherwise.

This can be made even more complicated. If $\sf ZF$ is inconsistent, then there is a model of $\sf ZF$ in the universe. To see why, note that from the meta-theoretic perspective, the universe is a model of $\sf ZF$, so $\sf ZF$ is consistent, but the internal interpretation is not. This means that there are non-standard integers which are the "source of the inconsistency" (either in inference rules, or in Replacement axioms, or in proof length).
Still, by the Reflection principle, every finitely many "true axioms" are consistent in some $V_\alpha$. So now we can ask what is the least $n$ so when we enumerate the axioms of $\sf ZF$ so that the "true axioms" appear first, such that there is no $V_\alpha$ which satisfies the first $n$ axioms of $\sf ZF$.
As we mentioned, this $n$ must be non-standard, so it is the case that there is some $V_\alpha$ which is a model of $\sf ZF$. Not internally, of course. From the internal point of view of the universe, it is only a model of a finite fragment of $\sf ZF$, but we know better. We know the truth. It is a model of all the true axioms of $\sf ZF$.
But this means that now going back into the universe, we have a least axiom of $\sf ZF$ such that the fragment below it is consistent, but the universe does not prove that it is a model of $\sf ZF$.
If you are willing to take this "sort of intermediate disagreement between the universe and its metatheory", then we can complete the circle:

*

*If $\sf ZF$ is at all consistent, take the least $L$-relation on $\omega$ which is a model of $\sf ZF$.


*If $\sf ZF$ is not consistent, then the least $\alpha$ such that $V_\alpha$ is a model of "the true axioms of $\sf ZF$", which we can identify in the universe.
In either case we get something that from our meta-theoretic point of view is a model of $\sf ZF$, but we cannot prove that it is a model of $\sf ZF$ from within the universe.
