TITLE: Are there ill-founded "maximally wide" models of $\mathsf{ZFC}$?
QUESTION [9 upvotes]: Throughout assume $\mathsf{ZFC}$ + "There is a proper class of inaccessible cardinals." I'm also happy to strengthen the large cardinal hypothesis if that would help.
Say that a model $M\models\mathsf{ZFC}$ is powerful iff every end extension satisfying $\mathsf{ZFC}$ is a top extension. The transitive powerful models are exactly the $V_\alpha$s where $\alpha$ is a worldly cardinal: that every $V_{\mathsf{worldly}}$ is powerful is trivial, and the large cardinal hypothesis above gives the other half of the result.
The ill-founded situation is on the other hand completely unclear to me:

Is there an ill-founded powerful model of $\mathsf{ZFC}$?

I don't really see where to start with this. Certainly there are no countable powerful models of $\mathsf{ZFC}$, ill-founded or otherwise, since we can (genuinely) force over such models; however, I don't see any useful tools for analyzing uncountable ill-founded models. The existence of rigid ill-founded models (e.g. the substructure of any ill-founded model of $\mathsf{ZFC+V=L}$ consisting of the parameter-freely-definable elements) does make the existence of powerful models feel not totally implausible, but that said I don't see how any of the techniques for building examples of the former type are useful for attempting to build an example of the latter.

REPLY [7 votes]: Back in the mid 1980s I remember convincing myself (alas, in unpublished work) that there is an ill-founded model $M$  of ZFC that has no end extension to another model of ZFC. Such a model $M$ by default is powerful and technically answers the question.
My unpublished work above used techniques employed in the following related results from the same time period:
1. There are ill-founded models $M$ of any given consistent extension of ZFC  that have no top extension to another model of ZFC;  see Theorem 1.5 of this paper of mine. Note that top extensions are referred to as rank extensions in the paper.
2. There are ill-founded models $M$ of any given consistent extension of ZFC such that there are no $N \supsetneq M$ such that $N$ is a forcing end extension of $M$. This was noted in this other paper of mine (see the remark after Theorem 1.5, which itself relates to Theorem 1.4).
In (1) and (2) above, the cardinality of $M$ can be arranged to be $\aleph_1$. This is optimum since every countable model of ZF (including ill-founded ones) has lots of top extensions and forcing extensions