This question reminds me of a magical little-known theorem of Jean Pierre Ressayre that shows that every nonstandard model of $PA$ has a model of $ZF$ as a submodel of its Ackermann intepretation, more specifically:
Theorem. [Ressayre] Suppose $(M, +, \cdot)$ is a nonstandard model of $PA$, and $\in_{Ack}$ is the Ackermann epsilon on $M$, i.e., $a\in_{Ack}b$ iff $\mathcal{M}$ satisfies "the $a$-th digit in the binary expansion of $b$ is 1". Then for every consistent recursive extension $T$ of $ZF$ there is a subset $A$ of $M$ such that $(A,\in_{Ack})$ is a model of $T$.
Proof Outline: By Löwenheim-Skolem, it suffices to consider the case when $M$ is countable. Choose a nonstandard integer $k$ in $M$, and consider the submodel $M_k$ of $(M,\in_{Ack})$ consisting of sets of ordinal rank less than $k$ [as computed within $(M,\in_{Ack})$]. "Usual arguments" show that $(M,\in_{Ack})$ has a Tarskian truth-definition for $M_k$, which in turn implies that $(M_k,\in_{Ack})$ is recursively saturated. Since $M_k$ is also countable, $(M_k,\in_{Ack})$ must be resplendent [which means that it has an expansion to every recursive $\Sigma^1_1$ theory that its elementary diagram is consistent with].
Now add a new unary predicate symbol $A$ to the language ${\in}$ of set theory and consider the (recursive) theory $T^A$ consisting of sentences of the form $\phi^A$, where $\phi \in T$, and $\phi^A$ is obtained by relativizing every quantifier of $\phi$ to $A$. It is not hard to show that $T^A$ is consistent with the the elementary diagram of $(M_k,\in_{Ack})$, so by replendence the desired $A$ can be produced.
[I will be glad to add clarifications]
Ressayre's theorem appears in the following paper:
J. P. Ressayre, Introduction aux modèles récursivement saturés, Séminaire Général de Logique 1983–1984 (Paris, 1983–1984), 53–72, Publ. Math. Univ. Paris VII, 27, Univ. Paris VII, Paris, 1986.
