Claim: If $X$ and $Y$ are equinumerous sets, then so are $2^X$ and $2^Y$.
Proof: Since $X$ and $Y$ are equinumerous, there exists a bijection $f: X\to Y$. With this, we can construct a function $F: 2^X\to 2^Y$ as follows:
$$U\subset X \mapsto f(U).$$
To show that $F$ is onto, let $V$ be any subset of $Y$. Construct a set $E$ as follows:
$$X\supset U = \{f^{-1}(v) : v\in V\}.$$
It's easy to see that $F(U) = V$. To show that $F$ is one-to-one, let $U, E\subset X$ be distinct and $\textbf{choose}$ some $x\in U\,\triangle\, E$. WLOG, assume that $x\in U$. Since $f$ is injective, then $f(x)\in F(U)$ but $f(x)\notin F(E)$.
There is no choice.
If $V \subset Y$, then $f(f^{-1}(V)) = V$. This is true for any surjective $f : X \to Y$.
If $f(U_1) = f(U_2)$, then $U_1 = f^{-1}(f(U_1)) = f^{-1}(f(U_2)) = U_2$. This is true for any injective $f$.