= Solution
Suppose $\operatorname{ad}(x)$ has a nonzero <eigenvalue> $\lambda$ and corresponding vector $y$. Then $[x,y]=\lambda y$, so
$$
\operatorname{ad}(y)x=-\lambda y\ne0,\qquad
(\operatorname{ad}y)^2x=0.
$$
For a <diagonalizable> <endomorphism> $T$, its <eigenspace> decomposition shows $\ker T^2=\ker T$. The displayed equations therefore contradict diagonalizability of $\operatorname{ad}(y)$.
Thus no $\operatorname{ad}(x)$ has a nonzero <eigenvalue>. Each is <diagonalizable> over the <algebraically closed field>, so each is the zero <endomorphism>. Hence $[x,z]=0$ for every $x,z\in\mathfrak g$, proving
$$
\boxed{\mathfrak g\text{ is abelian}.}
$$
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