I want to show that for $M\triangleleft R$.
$M$ is maximal $\iff$ $R/M$ is simple.
So first I made some observations:
$\begin{matrix}M\textrm{ is maximal } & & R/M \textrm{ is simple}\\\Updownarrow&&\\R/M\textrm{ is a field}\end{matrix}$
So firstly if $M$ is maximal, $R/M$ is a field, and if a unit element, $u$, gets into an ideal $I$, then since $rI\subseteq I, \quad\forall r\in R$, $r=u^{-1}$ means that $uu^{-1}\in I$ and then $1\in I$ meaning that $I=R$. If there are no unit elements in an ideal $I$, then $M=\{0\}$ since $R/M$ is a field. So any ideal in a field is $\{0\}$ or $R$ itself. Which means that $R/M$ being a field, means $R/M$ is simple.
$\begin{matrix}M\textrm{ is maximal } & & R/M \textrm{ is simple}\\\Updownarrow&\large{\nearrow}\!\!\!\!\!\!\large{\nearrow}&\\R/M\textrm{ is a field}\end{matrix}$
So we need now only get one arrow coming back from $R/M$ simple. This is what I am having trouble with. Some thoughts, perhaps the fact that $R/M$ is simple, and hence has no nontrivial ideals, means that we have no non-unit elements, hence is a field.
Is it possible to always make a new ideal somehow using nonunit elements?
If $M$ is maximal, $R/M$ is not necessarily a field. This is true when $R$ is commutative and has an identity. In fact, if $R$ is not commutative, then $R/M$ might not even be a division ring (the non-commutative version of a field).
Theorem. For any ideal $M$ in a ring $R$, there is a bijection between the set of ideals of $R$ which contain $M$ and the set of ideals of $R/M$, given by $I \mapsto I/M$.
So if $M$ is maximal, the only ideals of $R$ which contain $M$ are $R$ and $M$, which implies the only ideals of $R/M$ are $R/M$ and $M/M = \{M\}$ (the trivial ideal). Hence $R/M$ is simple.
Conversely, if $R/M$ is simple, its only ideals are $\{M\} = M/M$ and $R/M$, so the only ideals of $R$ containing $M$ are $R$ and $M$. Hence $M$ is maximal.