It seems to me that there do not exist two distinct idempotents matrix such that $(A-B)^2=0$. I have not found a counter-example with $2\times 2$ matrix
We can see that the hypotheses imply that $A+B=AB+BA$, and if we multiply by $AB$ on the right, we get $AB+BAB=(AB)^2+BAB$, which also implies that $AB$ is idempotent (and we can also see that $BA$ is idempotent).
After that, I am not sure what to do next."
Best Answer
Let $A=\begin{pmatrix}1 & 0 \\ 1 & 0\end{pmatrix}$, $B=\begin{pmatrix}1 & 0 \\ -1 & 0\end{pmatrix}$, so that $A-B=\begin{pmatrix}0 & 0 \\ 2 & 0\end{pmatrix}$.
Then $A^2=A, B^2=B$ and $(A-B)^2=0$.