If there exists a bijection between two sets $A$ and $B$, there exists a bijection between $\wp(A)$ and $\wp(B)$ Where $A$ and $B$ are countably infinite and $\wp(A)$ and $\wp(B)$  are their powersets.

$\#$ represents cardinality.

My proof is this:
Let $c = \#\wp(A) = 2^{\aleph_0}$.
If there exists a bijection between $A$ and $B$, then this implies (this is given in the definition I am using) that $\#A = \#B = \aleph_0$
Thus, $2^{\aleph_0} = 2^{\#A} = 2^{\#B} = \#(2^A)=\#(2^B)=c$

Thus, because both $2^A$ and $2^B$ have the same cardinality, by definition, this means that there exists a bijection between $2^A$ and $2^B$ or $\wp(A)$ and $\wp(B)$.

Is this enough? I proved that $2^{\#A} = \#(2^A)$ for finite sets, is that enough for countably infinite sets?
Let $f:A \to B$ be a bijection and define $F:\wp(A) \to \wp(B)$ by

$F(M):=f(M)$.

Then $F$ is a bijection.
Let $f:A \to B$ be a bijection and define $F:\wp(A) \to \wp(B)$ by

$F(M):=f(M)$.

Then $F$ is a bijection.