if $x$ and $y$ commute with $[x y]$, then $[x y]$ is nilpotent

Let $V$ be a complex-vector space and $x,y\in gl(V)$ such that $[x y]=xy-yx$ commutes with $x$ and $y$. Let $[x y]$. Prove that

(1)$tr(z^{m})=0$ , for every $m\geq 1$.

(2)Deduce that z is a nilpotent map.

I couldn't do the first one. For the second one if we can prove that the possible eigen values of $z$ is only $0$, we are done. From the first one , we have if $\lambda_{1},... \lambda_{n}$ are eigen values of $z$, then $\sum_{i=1}^{n} \lambda_{i}^{k} =0$, for every $k$. From there how can we comclude that $\lambda_{1}=\lambda_{2}=...= \lambda_{n}=0$.

Thanks in advance for any help.

Assuming that $z=[xy]$, since $z$ and $x$ commute we have $$ z^m=xyz^{m-1}-yxz^{m-1}=xyz^{m-1}-yz^{m-1}x$$ Hence $$ \mathrm{tr}(z^m)=\mathrm{tr}(xyz^{m-1})-\mathrm{tr}(y(z^{m-1}x))=\mathrm{tr}(xyz^{m-1})-\mathrm{tr}(xyz^{m-1})=0$$

For the second part, as you noted in your question it follows from the first part that $$ \sum_{k=1}^n\lambda_k^m=0$$ for all $m\geq 1$, where the $\lambda_k$ are the eigenvalues of $z$. It then follows from Newton's identities that the elementary symmetric polynomials $e_m(\lambda_1,\dots,\lambda_n)$ are zero for $m\geq 1$, so all the eigenvalues of $z$ are zero and $z$ is nilpotent.