Let $L$ be a simple group and $G=L^t$. The diagonal $D=\lbrace(x,x,\ldots,x), x\in L\rbrace$ is a subgroup of $L^t$. I can prove that if $D$ is maximal then $t$ is a prime.
If $t=mn$ then the subgroup $\lbrace (x_1,\ldots,x_1,x_2,\ldots,x_2,\ldots,x_n,\ldots,x_n), x_i\in L\rbrace$ properly contains the diagonal. Hence if $D$ is maximal then $t$ is a prime.
I can not prove the converse, namely if $D$ is a maximal subgroup of $L^p$ provided $p$ prime. For $p=2$ the statement holds.
EDIT
The problem that generated the previous question is the following: let $L$ be a simple group and $p$ be prime. The cyclic group with $p$ elements $\mathbb{Z}_p$ acts on $L^p$ by cycling the components, i.e.
$$a\cdot (x_1,x_2,\ldots, x_p)=(x_p,x_1,x_2,\ldots,x_{p-1})$$ where $a$ is a generator of the cyclic group. We can then build the semidirect product $G=L^p\rtimes \mathbb{Z}_p$ defined by the action above. The centralizer of $(1,1,\ldots,1,a)$ in $G$ is $$C=\lbrace(x,\ldots,x,a^n), x\in L, n\in \mathbb{N}\rbrace.$$
The question is: when is $C$ a maximal subgroup of $G$?
In particular, if the diagonal $D$ is maximal in $L^t$ then $C$ is maximal in $G$, but I am not sure that the converse hold (so the two questions are not equivalent).
Let $G = L^p \rtimes \mathbb Z_p$ (until the last paragraph it is actually not important that $p$ is prime). Suppose $\mathbb Z_p \le H \le G$. Then by the modular property of groups $$H = H \cap (L^p \mathbb Z_p) = (H \cap L^p) \mathbb Z_p.$$ Moreover $H \cap L^p$ is $\mathbb Z_p$-invariant. Conversely, for any $\mathbb Z_p$-invariant subgroup $S$ of $L^p$, the group $S \mathbb Z_p$ is a subgroup of $G$ satisfying $\mathbb Z_p \le S \le G$. Thus subgroups of $G$ containing $\mathbb Z_p$ are in natural bijection with $\mathbb Z_p$-invariant subgroups of $L^p$.
Now as you say the centralizer of $\mathbb Z_p$ is $C = D \mathbb Z_p$. To check that this is maximal it is sufficient to check that $D$ is a maximal among $\mathbb Z_p$-invariant proper subgroups of $L^p$.
Now suppose $D \le S \le L^p$. The (as guessed by Geoffrey Trang in the comments) one can show that there is an equivalence relation $\sim$ on $\{1, \dots, p\}$ so that $$S = \{(x_1, \dots, x_p) : i \sim j \implies x_i = x_j\}.$$ To see this, let $\Delta \subseteq \{1, \dots, p\}$ be a minimal nonempty set containing the support of some element of $S$. Let $i, j \in \Delta$ and suppose $x \in S$ satisfies $x_i \ne x_j$. Then $S$ contains an element $g$ such that $g_i = 1$ and $g_j \ne 1$. Since $L$ is simple, this implies that, for every $y \in L$, there is an element $g \in S$ such that $g_i = 1$ and $g_j = y$. Choose $y$ to be an element not commuting with $x_j$. Then $[x,g]$ is a nontrivial element of $S$ supported on $\Delta \setminus \{i\}$, a contradiction. Therefore $x_i = x_j$ for all $x \in S$ and $i, j \in \Delta$. Moreover, a similar argument shows that $S$ contains every element $x \in L^p$ supported on $\Delta$ such that $x_i = x_j$ for all $i, j \in \Delta$. Applying this to every choice of $\Delta$ gives the result.
Finally suppose $D \le S \le L^p$ and $S$ is $\mathbb Z_p$-invariant. Then $S = \{(x_1, \dots, x_p) : i \sim j \implies x_i = x_j\}$ for some $\mathbb Z_p$-invariant equivalence relation $\sim$. The blocks of the equivalence relation must all have the same size. Since $p$ is prime, this implies that there is either $1$ block or $p$, i.e., $D = S$ or $D = L^p$.
More generally, essentially the same argument shows that the diagonal group $D \cong L$ is maximal in $L^n \rtimes H$ if and only if $L$ is nonabelian simple and $H \le S_n$ is primitive or $n=2$ (see Dixon and Mortimer, Permutation Groups, Theorem 4.5A).