$A^2+B^2=AB$ and $BA-AB$ is non-singular The question is:

Are there square matrices $A,B$ over $\mathbb{C}$ s.t. $A^2+B^2=AB$ and $BA-AB$ is non-singular?

From $A^2+B^2=AB$ one could obtain $A^3+B^3=0$. Can we get something from this?
Edit: 
$$A^2+B^2=AB\implies\\
A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\
A^3+(A^2+B^2)B=A^2B \implies \\ 
A^3+A^2B+B^3=A^2B \implies \\
A^3+B^3=0
$$

We consider the equations
(1) $A^2+B^2=AB$,
(2) $A^2+B^2=2AB$.
A couple $(A,B)$, solution of (2), is simultaneously triangularizable, and, moreover, when $n=2$, $AB=BA$.
Eq. (1) has not the second property: indeed $$A=\begin{pmatrix}0&-1\\1&-1\end{pmatrix},\ B=\begin{pmatrix}\dfrac{1-i\sqrt{3}}{2}&0\\\dfrac{3+i\sqrt{3}}{2}&-1\end{pmatrix}$$ is a solution s.t. $AB\not= BA$. Note that $\mathrm{spectrum}(A)=\{e^{\pm 2i\pi/3}\}$ and $A^3=I_2$, $B^3=-I_2$; moreover $(AB-BA)^2=0$. I think that Eq. (1) has the first property, or at least, $AB-BA$ is nilpotent, but I don't know how to prove it. We see, in the following, that, a solution, in general position, satisfies $AB=BA$.
Proposition. We consider the equation in the unknown $X$: $X^2-AX+A^2=0$. Let $\mathrm{spectrum}(A)=(\lambda_i)_i$ denote the complete spectrum of $A$. If, for every $i\not= j$ $\lambda_i^3\not=\lambda_j^3$, then any solution $X$ satisfies $AX=XA$.
Proof. Since $A$ is diagonalizable we may assume $A=\mathrm{diag}(\lambda_i)$. Thus $A^3=\mathrm{diag}(\lambda_i^3)$ has distinct eigenvalues; since $X$ and $A^3$ commute, $X$ is a diagonal matrix and we are done.
EDIT. There are solutions of $A^3+B^3=0$ s.t. $AB-BA$ is invertible; for example, choose $A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\ B=\begin{pmatrix}0&0&0\\1&0&0\\0&-1&0\end{pmatrix}$; of course $A^2+B^2\not= AB$.
