Show that $\mathcal P(x)\notin x$ Does there exist a set $x$ such that $\mathcal P(x)\in x$, where $\mathcal P(x)$ denotes the power set of $x$?
In Zermelo-Fraenkel set theory, the axiom of regularity implies that for all sets $x$ and $y$, we cannot have $x\in y\in x$. Since $x\in\mathcal P(x)$, we cannot have $\mathcal P(x)\in x$, which gives an elementary proof.
However, many authors do not include the axiom of regularity in Zermelo-Fraenkly theory, mainly because almost all mathematical results can be proved without it, and the point of this axiom is basically a matter of convenience.
Therefore, I would like a proof which does not use the axiom of regularity either that we cannot have $\mathcal P(x)\in x$, or that this question is undecidable without this axiom.
In order to prove that this question is undecidable, I tried to show that the existence of a set $x$ such that $\mathcal P(x)\in x$ amounts to the existence of a set $a$ such that $a\in a$, or a set $b$ such that $b=\{b\}$, which (correct me if I'm wrong) are known to be two undecidable questions without the axiom of regularity.
 A: As Asaf stated, you cannot prove $\mathcal{P}(x)\notin x$ without the axiom of regularity. However, your strategy could not work for showing the desired independence: it implies the existence of a set $a$ such that $a\in a$, but the possibility that $\mathcal{P}(x)\in x$ leads to a contradiction (so it proves anything) is still opened.
Fortunately, Aczel's anti-foundation axiom, which is known to be consistent with ZFC without regularity, proves the existence of $x$ containing its power set as its element. The main consequence of Aczel's axiom is the Solution lemma:

Solution Lemma. Working with sets with a class $X$, possibly proper, of atoms. If $a_x$ is a $X$-set (i.e. a set which is a mixture of other sets and atoms of $X$) then the system of equations
  $$x = a_x \quad(x\in X)$$
  has a unique solution $(b_x\mid x\in X)$; i.e. the equality $x=a_x$ still holds if we hereditarily replace $x$ to $b_x$

(see Aczel's Non-well-founded set theory for its details.)
Consider the following system of equations for $y$:
$$
y = \mathcal{P}(\{y,\varnothing\}).
$$
We can see the Solution Lemma attests the existence of sets $y$, and moreover, $x=\{y,\varnothing\}$ satisfies $\mathcal{P}(x)\in x$.
