= Solution
There are two ways for a <set> to fail to be the <power set> of $y$: it may contain an object which is not a <subset> of $y$, or omit a <subset> of $y$. A single existential <logical formula> expressing these alternatives is
$$
\boxed{\phi(x,y)\equiv\exists z\Bigl[
\bigl(z\in x\wedge(\exists u\in z)\,u\notin y\bigr)
\vee\bigl(z\notin x\wedge(\forall u\in z)\,u\in y\bigr)\Bigr].}
$$
Its matrix is bounded, so this is $\Sigma_1$. Its negation is precisely the first-order power-set specification
$$
\forall z\bigl(z\in x\ \longleftrightarrow\ (\forall u\in z)\,u\in y\bigr).
$$
We prove the absoluteness fact needed to exclude a logically equivalent $\Sigma_1$ <logical formula>. If $M\subseteq N$ are <transitive sets> and all parameters lie in $M$, bounded <logical formulas> have the same truth value in $\langle M,\in\rangle$ and $\langle N,\in\rangle$. Atomic equality and membership agree because both structures use actual membership. Boolean operations preserve agreement. In a bounded <quantifier> over $v\in M$, transitivity ensures that every element of $v$ already lies in $M$, so the two structures quantify over exactly the same elements of $v$. Induction on <logical formulas> proves the claim. It follows that $\Sigma_1$ <logical formulas> are upward absolute: witnesses in $M$ remain available in $N$, and their bounded matrix retains its truth value.
Now use a finite counterexample, so no existence of a transitive model of <ZF> is assumed. Let $M=4=\{0,1,2,3\}$, the <Finite von Neumann ordinal>, and let $N=V_4$. These are <transitive sets> with $M\subseteq N$. <Set> $y=2$ and $x=3$, both elements of $M$. Inside $M$, the <subsets> of $2$ which are available as elements of the domain are exactly $0,1,2$; the remaining domain element $3$ is not a <subset> of $2$. Thus $M$ satisfies the displayed power-set specification with $x=3$.
Inside $N$, however, the <set> $\{1\}$ is also present and is a <subset> of $2$, while $\{1\}\notin3$. Thus the same specification is false in $N$. If $\neg\phi$ were logically equivalent to a $\Sigma_1$ <logical formula>, its truth in $M$ would transfer upward to $N$, a contradiction. Therefore
$$
\boxed{\neg\phi\text{ is not logically equivalent to any }\Sigma_1\text{ formula}.}
$$
This proves the requested unrestricted logical non-equivalence. The finite structures are not being claimed to satisfy all of <ZF>; logical equivalence must hold in them as well as in any models of a stronger theory.
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