Let $G$ be a finite nonabelian simple group. Must $G$ admit a pair of generators $g,h$ such that $g,[g,h]$ also generate?
I've computationally verified this for the first 21 nonabelian finite simple groups, though I don't know if this should hold in general.
Yes, this is true, at least for large enough groups. It is known that a random pair of elements generate a finite simple group (with probability approaching $1,$ as size goes to infinity), and also
Robert M. Guralnick, Martin W. Liebeck, Jan Saxl, and Aner Shalev, MR 1707675 Random generation of finite simple groups, J. Algebra 219 (1999), no. 1, 345--355.
Show that for random $x, y,$ the elements $x$ and $x^y$ generate with probability again approaching $1.$ as size goes to infinity.
The combination of the last two statements shows that your conjecture is true for large groups. It's quite possible that you can extract the statement for all simple groups by reading the referenced paper.
Here is another approach. As noted in the comments, the question is equivalent to asking which finite simple groups can be generated by two conjugate elements. It is known that most, but not all, non-Abelian simple groups can be generated by an element of order $2$ and an element of order $3$. It is proved in https://arxiv.org/abs/1603.04717 by Carlisle King that every finite simple group can be generated by two elements of prime order, and that one of the elements may be chosen to have order two. However if $G$ is a non-Abelian finite simple group with $G = \langle x,t \rangle,$ where $t$ has order $2$, then $\langle x,t^{-1}xt \rangle$ is normalized by both $x$ and $t$, so is normal in $G,$ and is all of $G$. Hence every non-Abelian finite simple group can be generated by a pair of conjugate elements.
