**EDIT :** The first version of my answer was unnecessarily complicated. The following avoids lifting of idempotents and structure theory of artinian rings and answers the original question:

So let $R$ be a ring with $1$ and finite maximal ideal $M$. (Note that if $R$ contains a finite maximal left ideal, then it contains a finite maximal ideal.) Let

$$ I = \{ r\in R \mid rM = 0 \}. $$
I claim that if $R/M$ is infinite, then $R = I \oplus M \cong (R/M)\times (R/I)$. Thus the answer in the noncommutative case is: **finite ring or direct product of simple ring and finite ring.**

Proof of $R=I+M$: Note that $I$ is the kernel of the map
$$ R \to \bigoplus_{m\in M} (Rm)\subseteq M^M,\quad r\mapsto (rm)_{m\in M}. $$
As $M$ is finite, $R/I$ is finite. Since we assume $R/M$ infinite, it follows that $I+M> M$ and so $I+M=R$ by maximality.

Proof of $I\cap M=0$: Assume $I\cap M\neq 0$. Since $M$ is finite, $(I\cap M)_R$ contains a simple right $R$-module $S_R$. As $IM=0$, we may view $S$ as a module over $R/M$. Let $D= \text{End}(S_R)$. Since $|S|$ is finite, it follows $S\cong D^n$ for some $n$ and $D$ is finite. By the density theorem of Jacobson-Chevalley then $R/M$ is isomorphic to an $n\times n$-matrix ring over $D$ and thus finite, contradiction. Thus $I\cap M=0$ as claimed.

In the special case where $R$ has a finite maximal left ideal, an occuring infinite simple ring factor is necessarily a division ring.