I think the answer to your question is yes, even for higher powers but let me stick to squares.
Every nilpotent matrix is conjugate to a nilpotent matrix in Jordan form, which is unique up to permutation of Jordam blocks. So we have a bijection
$$\mathrm{Nilp}_n(\mathbb C)/\mathrm{conjugation} \quad \cong \quad \mbox{integer partitions of }n$$
associating with a conjugacy class of a nilpotent matrix $X$ the sizes of its Jordan blocks $(a_1, \ldots, a_r)$ which sum up to $n$. The max of the $a_i$'s is the nilpotency-degree of $X$. To the class of $X^2$ is associated the partition $(a_1-1,1,a_2-1,1, \ldots, a_r-1,1)$. From here you can derive a necessary and sufficient condition on a nilpotent matrix to be a square.
Now fix your preferred nilpotent matrix $N$. Let $X$ be a matrix with $X^2=N$ and Jordan type $(a_1, \ldots, a_r)$ where $r$ is minimal. Conjugating the whole setup, we may assume $X$ is in Jordan form.
Let $Y$ be a matrix with $Y^2=X^2 = N$ and let us construct a path from $X$ to $Y$. Let $(b_1, \ldots, b_s)$ be the Jordan type of $Y$. By minimality of $r$ we have $b_i=a_i$ whenever $b_i>1$, and if $b_i=1$ we have $a_i=1$ or $a_i=2$. Look the $2\times 2$ Jordan blocks of $X$: Replacing the $1$ in appropriately many of those blocks by a $t$ and varying $t$ from $1$ to zero yields a path inside the set of "square roots" $\mathcal S_N$ from $X$ to a matrix $Z$ in Jordan form having the same Jordan type as $Y$.
It remains to construct a path from $Z$ to $Y$ inside $\mathcal S_N$. Since $Z$ and $Y$ have the same Jordan type, there exists an invertible matrix $S$ with $Y=SZS^{-1}$. Because $Y^2=Z^2=N$ the matrix $S$ commutes with $N$. It is enough to construct a path from the identity matrix to $S$ in the set $\mathcal C_N$ of invertible matrices that commute with $N$.
I claim $\mathcal C_N$ is path connected. Indeed, the set $[N]$ of all commutators of $N$ is linear subspace of the vector space $\mathrm M_n(\mathbb C)$. The determinant, as a function on $[N]$ is a polynomial function which is not identically zero since the identity matrix belongs to $[N]$. Thus, the zero set of the determinant is Zariski closed, so $\mathcal C_N$ is Zariski open in $[N]$. Any Zariski-open in a complex vector space is path connected.