Does $2^X \cong 2^Y$ imply $X \cong Y$ without assuming the axiom of choice?

In fact, not only is it not provable without the Axiom of Choice, it's not provable with the Axiom of Choice! For instance, it's consistent with AC that $2^{\aleph_0} = 2^{\aleph_1} = \aleph_2$. On the other hand, if $X$ and $Y$ are finite then certainly $2^X\cong 2^Y \implies X\cong Y$, and proving this doesn't require AC at all since all the quantities involved are finite. More broadly, Easton's Theorem says that aside from some mostly-trivial constraints (e.g., $A \gt B \implies 2^A \gt 2^B$), the cardinalities of power sets (of regular cardinals) can be entirely arbitrary.

Actually this statement is not even provable within ZFC.

See this mathoverflow question