Let $G$ be a direct product of nonabelian simple groups $T_1,T_2,\dots,T_d$ with $d>1$. Can $G$ be generated by two elements of $G$?
This answer mostly summarizes what was written in the comments:
(1) If the $T_i$ are pairwise nonisomorphic, this will happen. Each of the $T_i$ will individually be $2$-generated (see here; this uses the classification of finite simple groups). Now, let $(g_i, h_i)$ be a pair of generators for $T_i$ and let $g = (g_1, \ldots, g_d)$ and $h = (h_1, h_2, \ldots, h_d)$; I claim that $g$ and $h$ generate $\prod T_i$. Indeed, let $G = \langle g,h \rangle$. Then $G$ surjects onto each $T_i$, so it has each $T_i$ as a Jordan-Holder factor. Using that the $T_i$ are pairwise nonisomorphic, $G = \prod T_i$.
(2) At the other extreme, for a fixed finite simple group $T$, the number of generators needed to generate $T^d$ goes to $\infty$ as $d \to \infty$; see here.