AB diagonalizable then BA also diagonalizable

Try $A = \begin{bmatrix}0&0\\1&0\end{bmatrix}$ and $B = \begin{bmatrix}0&0\\0&1\end{bmatrix}$

This is true in the special case when both $A$ and $B$ are invertible.

If we denote by $E_C(\lambda)$ the eigenspace of the matrix $C$ relative to the eigenvalue $\lambda$, we can see that $$ v\mapsto Bv $$ induces an injective linear map $E_{AB}(\lambda)\to E_{BA}(\lambda)$, so $\dim E_{AB}(\lambda)\le\dim E_{BA}(\lambda)$, where $\lambda$ is any eigenvalue of $AB$. By symmetry, the two eigenspaces have the same dimension.

Also this proves that $AB$ and $BA$ have the same eigenvalues. If $AB$ is diagonalizable, then the geometric multiplicities of its eigenvalues sum up to $n$ (the size of the matrices $A$ and $B$). Thus the same happens for $BA$ and $BA$ is diagonalizable.

If $A$ or $B$ is not invertible, we can still see that $AB$ and $BA$ have the same nonzero eigenvalues. Indeed, if $\lambda$ is a nonzero eigenvalue of $AB$, with eigenvector $v$, then $ABv=\lambda v$ (which implies $Bv\ne0$, so $(BA)(Bv)=\lambda(Bv)$ and therefore $\lambda$ is an eigenvalue of $BA$. By symmetry, $AB$ and $BA$ share the nonzero eigenvalues. Also, $0$ is an eigenvalue of both, because neither is invertible. However, in this case we can't control the geometric multiplicities, as Michael Biro's example shows.