Similarity of squared matrix products with one diagonalizable product ($AB\text{ diagonalizable}⟹(AB{)}^2\sim (BA{)}^2$)

The identity $(BA{)}^{j+1}=B(AB{)}^{j}A$ gives $m_{BA}(t)\mid t{m}_{AB}(t)$. If $AB$ is diagonalizable, its minimal polynomial has simple roots. Hence all nonzero Jordan blocks of $BA$ have size one and its zero blocks have size at most two. Squaring kills these nilpotent blocks, making both squared products diagonalizable. Their eigenvalue multisets agree by Characteristic polynomials of AB and BA, so the squared products are similar matrices.