Does the Zariski closure of a maximal subgroup remain maximal?

Let $k$ be an algebraically closed field and let $G\leq\rm{GL}_n(k)$. Assume that $M<G$ is a maximal subgroup (in the abstract group sense). Denote by $\bar{G}^Z$ the Zariski closure of $G$ in $\rm{GL}_n(k)$. Is it true that $\bar{M}^Z<\bar{G}^Z$ is a maximal subgroup in the algebraic groups sense? If yes, would it be a maximal subgroup in the abstract group sense?
Thanks in advance for any help.

The following is a different type of a counterexample. Not sure that it qualifies or matches with what you were looking for. Let $$ G=\left\{\left(\begin{array}{cc}1&m\\0&1\end{array}\right)\mid m\in\Bbb{Z}\right\}. $$ We view this as a subgroup of $GL_2(\Bbb{C})$. It is isomorphic to the additive group of integers, so the subgroup $M=M_p$ consisting of the elements of $G$ such that $m$ is divisible by a fixed prime $p$ is a maximal subgroup.

But both $G$ and $M$ have as their Zariski closure the group $$ \overline{G}=\left\{\left(\begin{array}{cc}1&m\\0&1\end{array}\right)\mid m\in\Bbb{C}\right\}=\overline{M}. $$ So $\overline{M}$ is not a maximal subgroup of $\overline{G}$ in any sense.