Let $A,B\in M_n(\mathbb{C})$ such that $(A-B)^2=A-B$. Prove that $$\mbox{rank}(A^2 - B^2) \geq \mbox{rank}(AB - BA)$$
So far, for convenience, I have denoted
$$\begin{aligned} A-B &:=X\\ A+B &:= Y\\ A^2-B^2 &:=C\\ AB-BA &:=D\end{aligned}$$
Then we have
$$C-D=YX, \qquad C+D=XY,$$
therefore it remains to be proved that $rank(XY+YX)\geq rank(XY-YX)$ under the hypothesis $X^2=X$.
Let us get our hands dirty.
By the given condition, $P:=A-B$ is idempotent. By a change of basis, we may assume that $P=\pmatrix{I_r&0\\ 0&0}$ where $r=\operatorname{rank}(P)$. Partition $B$ accordingly as $\pmatrix{X&Y\\ Z&W}$, where $X$ is $r\times r$. Then \begin{aligned} A^2-B^2&=(B+P)^2-B^2=BP+PB+P^2=\pmatrix{2X+I_r&Y\\ Z&0},\\ AB-BA&=PB-BP=\pmatrix{0&Y\\ -Z&0}. \end{aligned} You may continue from here.