There isn't any maximal finite subgroup of a multiplicative group of a division ring. It is an exercise in division algebras and I couldn't find the answer:

Suppose that $D$ is a division ring. Prove that $D^{\times}$ has no finite maximal subgroup.

 A: Let $D$ be a division ring. If $D$ is finite, then by Wedderburn's Theorem, $D$ is a field and then $D^\times$ is cyclic and has a maximal subgroup (take the non-generating elements, since these are finite, one among them must be maximal). So the statement does not hold. In the case where $D$ is infinite, I have no found a coherent method that would solve the problem. After doing some reading, this seems to be an open research question. See the following papers for more:
ON MAXIMAL SUBGROUPS OF THE MULTIPLICATIVE GROUP OF A
DIVISION ALGEBRA

ON THE MULTIPLICATIVE GROUP OF A DIVISION
RING
SUBGROUPS OF THE MULTIPLICATIVE GROUP
OF A DIVISION RING
A: This is not answer but I state following easy lemma. It may help us.

Let $G$ be group and $M$ be a maximal subgroup of $G$. Then either $Z(G)\subseteq M$ or $G'\subseteq M$.

Now, if $(D^*)' \subseteq M$, then $M$ is a normal subgroup. Hence, there exists a prime number $p$ such that $[D^*:M]=p$. Since $M$ is finite, $D$ is finite too.
So we can assume that $M$ contains center of $D$. If $D$ is a finite dimension algebra on its center, then we can conclude that $D$ is finite.
