Yes, you can prove this in ZF. One way of seeing this is to note that $2^{\aleph_0}$ is the size of $A=\{0,1\}^{\mathbb N}$ (the set of functions from ${\mathbb N}$ to $\{0,1\}$), and $A\times A$ is easily seen to be in bijection with $\{0,1\}^{\mathbb N\sqcup\mathbb N}$, where $\sqcup$ denote disjoint union. But $\mathbb N\sqcup\mathbb N$ is in bijection with $\mathbb N$ (think even and odd numbers).

In fact, Cantor's classical proofs that $\mathbb R^2$ and $\mathbb R$ are in bijection do not use choice. For example, $\mathbb R$ and $(0,1)$ are easily seen to be in bijection (think $\arctan$ or somesuch), and one can find a bijection between $(0,1)^2$ and $(0,1)$ by looking at decimal expansions and intertwining. (Usually one needs to treat a small (countable) set a bit differently in these arguments.)

In general, the cardinal $2^\kappa$ is the size of the set of functions from a set of size $\kappa$ to $\{0,1\}$, so $2^\kappa\times 2^\kappa$ is $2^{\kappa+\kappa}$ (as with exponentiation of finite numbers), where the sum denotes the size of a disjoint union of two sets of size $\kappa$. So, it is enough to know that $\kappa+\kappa=\kappa$ to conclude that $2^\kappa\times 2^\kappa =2^\kappa$.

It is consistent with ZF to have infinite sets $A$ such that $A\times A$ is not in bijection with $A$. In fact, as already pointed out in Martin's answer, if there are no such exceptions, choice holds. It is also consistent to have sets $A$ such that $A\sqcup A$ and $A$ are in bijection, but $A\times A$ and $A$ are not. In that case, ${\mathcal P}(A)$ and ${\mathcal P}(A)\times{\mathcal P}(A)$ would be in bijection, as explained above.

Note that if $A$ is infinite and $A\times A$ and $A$ are in bijection, then so are $A\sqcup A$ and $A$ (by Schröder-Bernstein, which does not need choice). For the case you are asking, already $\mathbb N\times\mathbb N$ is in bijection with $\mathbb N$, so things work out nicely.