Is it possible to find $n\times n$ matrices $M,N$ such that $(MN)^n=0$ but $(NM)^n\ne 0$?
I can see that $(MN)^n=0\implies (NM)^{n+1}= 0$ by associativity. But I don't see that it is necessarily true for $(NM)^n$.
I can also see that $MN=0\not\implies NM=0$ but I cannot find an example for $(MN)^n=0$ but $(NM)^n\ne 0$.
On
You can combine your observation that $(MN)^n=0\implies (NM)^{n+1}=0$ with the result of this other question, showing that $(NM)^{n+1}=0\implies (NM)^n=0$.
No. As it turns out, $MN$ and $NM$ must have the same nonzero eigenvalues (although the multiplicities need not agree), so if $(MN)^n = 0$ then $MN$ has only zero eigenvalues, so the same must be true of $NM$, which must therefore have characteristic polynomial $t^n$.
One elegant proof uses the following lemma: in any ring, $1 - xy$ is invertible if and only if $1 - yx$ is invertible. Now use the fact that $\lambda$ is a nonzero eigenvalue of $MN$ if and only if $1 - \frac{MN}{\lambda}$ is not invertible.
Another proof proceeds as follows: let $M, N$ have entries $m_{ij}, n_{ij}$ which are formal variables in a polynomial ring $R = \mathbb{Z}[m_{ij}, n_{ij}]$. Observe that $$\det(Nt - NMN) = \det(N) \det(t - MN) = \det(t - NM) \det(N).$$
Since $R$ is an integral domain, we conclude that $$\det(t - MN) = \det(t - NM)$$
as polynomials in $R$. So in fact $MN$ and $NM$ must have the same characteristic polynomial. (Note, however, that the first proof is much more general; among other things, it remains valid in infinitely many dimensions.)