We prove that there exist such matrices $A$ and $B$ if and only if $n$ is even. I. Assume that $n$ is odd and some invertible $n \times n$ matrices $A, B$ satisfy $A B-B A=B^{2} A$. Hence $B=A^{-1}\left(B^{2}+B\right) A$, so the matrices $B$ and $B^{2}+B$ are similar and therefore have the same eigenvalues. Since $n$ is odd, the matrix $B$ has a real eigenvalue, denote it by $\lambda_{1}$. Therefore $\lambda_{2}:=\lambda_{1}^{2}+\lambda_{1}$ is an eigenvalue of $B^{2}+B$, hence an eigenvalue of $B$. Similarly, $\lambda_{3}:=\lambda_{2}^{2}+\lambda_{2}$ is an eigenvalue of $B^{2}+B$, hence an eigenvalue of $B$. Repeating this process and taking into account that the number of eigenvalues of $B$ is finite we will get there exist numbers $k \leq l$ so that $\lambda_{l+1}=\lambda_{k}$. Hence $$\lambda_{k+1} =\lambda_{k}^{2}+\lambda_{k} \ldots \lambda_{l} =\lambda_{l-1}^{2}+\lambda_{l-1} \lambda_{k} =\lambda_{l}^{2}+\lambda_{l}$$ Adding these equations we get $\lambda_{k}^{2}+\lambda_{k+1}^{2}+\ldots+\lambda_{l}^{2}=0$. Taking into account that all $\lambda_{i}$ 's are real (as $\lambda_{1}$ is real), we have $\lambda_{k}=\ldots=\lambda_{l}=0$, which implies that $B$ is not invertible, contradiction. II. Now we construct such matrices $A, B$ for even $n$. Let $A_{2}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ and $B_{2}=\left[\begin{array}{cc}-1 & 1 \\ -1 & -1\end{array}\right]$. It is easy to check that the matrices $A_{2}, B_{2}$ are invertible and satisfy the condition. For $n=2 k$ the $n \times n$ block matrices $$A=\left[\begin{array}{cccc} A_{2} & 0 & \ldots & 0 \\ 0 & A_{2} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & A_{2} \end{array}\right], \quad B=\left[\begin{array}{cccc} B_{2} & 0 & \ldots & 0 \\ 0 & B_{2} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & B_{2} \end{array}\right]$$ are also invertible and satisfy the condition.