I found out that the equation $(AB-BA)^m=I_n$ does have solution when $n=km$, where $k$ is an arbitrary integer number.
To prove, we just need to consider $C=\mathrm{diag}(r_1,..., r_m)$ where $r_1,..., r_m$ are the roots of $x^m-1=0$, i.e., the eigenvalues of matrix $C$.
In this case we know $\mathrm{trace}(C)=0$, and so there are two matrices $A$ and $B$ such that $C=AB-BA$, and also it is vivid that $C^m=I_m$. For each $n=km$, we can duplicate the matrix $C$, $k$ times to get an $mk\times mk$ matrix to consider as a new matrix $C$.
Now, my question is this:
Does the equation $(AB-BA)^m=I_n$ have answer if and only if $n=mk$?
As Alex Zorn wrote, let $m$ be a prime and assume that $\mathrm{char}(K)=0$; $x^m-1=(x-1)f_m(x)$ where $f_m$ is irreducible over $\mathbb{Q}$. Let $1,s_2,\dots,s_{m}$ be the roots of $x^m-1$. Note that $$\mathrm{spectrum}(C)=\{1,\alpha_1;s_2,\alpha_2; \dots;s_{m},\alpha_{m}\},$$ where the $(\alpha_i)_i$ are integers, $\alpha_2s_2+\cdots +\alpha_ms_m=-\alpha_1$ and $\sum_{i=1}^m\alpha_i=n$. Since $s_2,\dots,s_m$ is a free system over $\mathbb{Q}$, the relations $$-1=s_2+\dots+s_m=(\alpha_2/\alpha_1)s_2+\dots+(\alpha_m/\alpha_1)s_m$$ imply $\alpha_1=\dots=\alpha_m.$
Then the characteristic polynomial of $C=AB-BA$ is $\chi_C(x)=(x^m-1)^{\alpha_1}$; therefore $n=\alpha_1m$.