 Solution

Take the formula saying that a set is transitive:This is a bounded formula in set theory. If $x\in M$, transitivity of $M$ ensures that every $u\in x$ and every $v\in u$ is already in $M$, and the same holds in $N$. Both structures therefore run through exactly the same bounded witnesses and use the same actual membership relation. They agree on ${\phi }_1(x)$ for every $x\in M$, proving set-theoretic absoluteness.