As in Geoff Robinson's answer, let us show that there exists a square root of $A$ which is a polynomial in $A$; then surely it commutes with $B$.

Let $\mu(x)$ be the minimal annihilating polynomial for $A$, $\mu(x)=\prod_i(x-\lambda_i)^{\alpha_i}$. We need to find a polynomial $P(x)$ such that $P(x)^2\equiv x\pmod {\mu(x)}$ (then $P(A)$ is a desired square root). By the Chinese remainder theorem, it suffices to find such polynomial modulo each of $(x-\lambda_i)^{\alpha_i}$, which is the same as finding a square root of $t+\lambda_i$ modulo $t^{\alpha_i}$. This last root can be proved to exist either by simple lifting of exponent $\alpha_i$, or by mentioning that the Taylor polynomial
$$
P_i(t)=\sqrt{\lambda_i}\sum_{n=0}^{\alpha_i-1}
\frac{\frac12\bigl(\frac12-1\bigr)\cdots\bigl(\frac12-n+1\bigr)}{n!}\biggl(\frac{t}{\lambda_i}\biggr)^n
$$
for $(t+\lambda_i)^{1/2}$ fits.

This proof works as well if $\mu(x)$ is divisible by $x$ (but not by $x^2$) since $0$ is a square root of $x$ modulo $x$.

**EDIT.** As Geoff Robinson mentioned, my last statement was wrong, so the case when $\mu(x)$ is divisible by $x^2$ needs another methods. In this case, the desired claim is wrong. One may clearly use the statement that each operator commuting with all operators commuting with $A$ is a polynomial in $A$, but here is an explicit counterexample (surely inspired by Geoff).

Let $A=\pmatrix{0&0&1\\0&0&0\\0&0&0}$, $B=\pmatrix{1&0&0\\0&0&0\\0&0&1}$. Then $AB=BA$, and there exists a square root of $A$, for instance, $\pmatrix{0&1&0\\0&0&1\\0&0&0}$. On the other hand, since $B$ is the projector onto $\langle e_1,e_3\rangle$, each $C$ commuting with $B$ should have the form $C=\pmatrix{a&0&b\\0&c&0\\d&0&e}$. Now, if $C^2=A$ then $\pmatrix{a&b\\d&e}^2=\pmatrix{0&1\\0&0}$ which is impossible: otherwise we would obtain a $2\times 2$ matrix whose square is nonzero but the fourth power is zero.