Tarski's theorem, as given in Undecidable theories, page 46, allows arbitrary numbering and is completely syntactic. I think this abstract version given by Tarski himself is the most clear. Let me summarize it here with some inessential variations.

Let $T$ be a consistent first-order theory (any consistent first-order theory). If $\varphi\mapsto \ulcorner\varphi\urcorner$ is a naming of formulas (**any** assignment of closed terms to formulas), then either the diagonalization function (the function $\varphi\mapsto \varphi(\ulcorner\varphi\urcorner)$) is not representable (under that naming), or the set of theorems is not representable (under the given naming), or both are not representable.

In the case of ZF, assuming it to be consistent, we know that if we choose a recursive naming, we can represent the diagonalization function but not the set of theorems. Also, we can easily choose a (nonrecursive) naming which allows us to represent the set of theorems, but, then, the diagonalization will not be representable.

The proof is quite simple. If the diagonalization is representable, the fixed-point lemma can be proved quite simply. Assume that $V$ is a formula representing the set of theorems. Apply the fixed point lemma to get $\varphi$, a sentence satisfying
$T\vdash\varphi\leftrightarrow \neg V(\ulcorner\varphi\urcorner)$.

If $T\vdash\varphi$, then, since $V$ represents the theorems, $T\vdash V(\ulcorner\varphi\urcorner)$, and $T$ is inconsistent. If $T\nvdash \varphi$, then, since $V$ represents the theorems, $T\vdash\neg V(\ulcorner\varphi\urcorner)$, and $T\vdash \varphi$ by the choice of $\varphi$. Therefore, $T\vdash \varphi$ and it is inconsistent by the previous argument.

**EDIT**

Motivated by the question in the comment, I will prove the fixed point lemma I have used above:

We are assuming that $T$ is a first-order theory and that the diagonalization is represented in $T$ under the arbitrary naming $\varphi\mapsto\ulcorner\varphi\urcorner$. It means that there is a formula $D(x,y)$ such that
$T\vdash\forall y(D(\ulcorner\phi\urcorner, y)\leftrightarrow y=\ulcorner\phi(\ulcorner\phi\urcorner)\urcorner)$.

Now, let $W(y)$ be an arbitrary formula. Let $\phi(x)$ be the formula $\exists y(D(x,y)\wedge W(y))$ and let $\varphi$ be the sentence $\phi(\ulcorner\phi\urcorner)$, the diagonalization of $\phi$. This sentence is a fixed point for $W(y)$.

Indeed, $\varphi$ is $\exists y(D(\ulcorner\phi\urcorner,y)\wedge W(y))$, which, from the hypothesis on the representation of the diagonalization, is equivalent to $\exists y(y=\ulcorner\varphi\urcorner\wedge W(y))$. The last sentence is logically equivalent to $W(\ulcorner\varphi\urcorner)$, and we are done.

Therefore, Tarski's result applies to arbitrary first-order theories and to arbitrary namings. The moral is that no matter what first-order theory and naming of formulas you choose, the representation of at least one of two simple metatheoretical notions
(diagonalization and theoremhood) within the object theory will always fail.

isa theorem about models. It is about undefinability of "truth" which means satisfaction. We are showing that for any given model $M$ of ZFC (although this holds in more generality) is no formula $\phi(v)$ in the language of set theory such that $M \models \phi(x)$ iff $x$ is a standard natural number coding a sentence that is true in $M$. If you like, we can restrict our attention to one preferred model, say "V". I don't know how to phrase this result as a purely syntactic / provability result-- it's about satisfaction. $\endgroup$ – Monroe Eskew Apr 27 '20 at 19:34