If $N^n=0$ but $N^{n-1}\neq 0$, then there is no $n\times n$ matrix $A$ such that $A^2=N$
Solution 1:
It is assumed that $n \neq 1$. $A^{2n}=0$ and this implies that $A^{n}=0$ (because $A$ is $n \times n$). Hence, $A^{k}=0$ for all $k \geq n$. But $2n-2 \geq n$ so $A^{2n-2}=0$. So $N^{n-1}=0$, a contradiction.
[To show tat $A^{n}=0$ observe that $0$ is the only eigne value. Use the fact that $p(A)=0$ where $p$ is the characteristic polynomial (Cayley Hamilton Theorem:
https://en.m.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem )].