= Solution
To prove that <power-set equality is not upward absolute>, use the standard <constructible universe theorem>: $L$ is a transitive <inner model> of <ZF> containing every <ordinal>, and <set> rank is absolute between $L$ and $V$. Since $V\ne L$, some rank stage contains a nonconstructible <set>. Let $\alpha$ be the least <ordinal> with $V_\alpha\nsubseteq L$. It is not zero; it cannot be a limit, since a <union> of earlier stages contained in $L$ is still contained in $L$. Thus $\alpha=\beta+1$ and $V_\beta\subseteq L$.
We need not merely this inclusion, but that $y=V_\beta$ itself belongs to $L$. Inside $L$, <ZF> constructs its own rank stage $(V_\beta)^L$, which is a <set> of $L$. By transitivity and <set-theoretic absoluteness>[rank absoluteness],
$$
(V_\beta)^L=V_\beta\cap L=V_\beta.
$$
Consequently $y\in L$. Let $x=(\mathcal P(y))^L$, the <power set> computed in $L$. Then $x\in L$ and $L\models x=\mathcal P(y)$.
By minimality there is $z\in V_{\beta+1}\setminus L$. Externally $z\subseteq V_\beta=y$, but $z\notin x$, because every member of the <set> $x\in L$ is in the <transitive class> $L$. Hence $V\not\models x=\mathcal P(y)$. Taking $U_1=L$ and $U_2=V$ proves \b[<power-set equality is not upward absolute>, even between transitive models of <ZF>].
There is no contradiction with absoluteness of $z\subseteq y$: that bounded relation agrees whenever both parameters are present in the smaller class. <power-set equality is not upward absolute>[Power-set equality] quantifies over all <subsets>, including a new $z$ absent from the smaller class. The step proving $V_\beta\in L$ is essential; inclusion in a class alone would not imply membership in it.
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