Maximal ideals are prime ideals in the non-commutative case

Let $R$ be a ring with unity, not necessarily commutative. Show that if $M$ is a maximal two-sided ideal in $R$, then $M$ is prime.

I know several proofs in the commutative case, but I can not come up with one in the general case.

Let $I,J$ be two ideals such that $IJ\subset M$. Assume $I\not\subset M$. Then, there is $a\in I$ such that $R=M+(a)$ as two-sided ideals because $M$ is maximal. In the commutative case it is easy to proceed from here, but here I just get that $1=m+ras$ for some $m\in M, r,s\in R$ and multiplying with an element $b\in J$ doesn't give anything?

We shall prove it by contradiction. Let $I,J$ be two-sided ideals such that $IJ\subseteq M$ but $I\not\subseteq M$ and $J\not\subseteq M$. Then $M\subsetneq M+I$ because $I\not\subseteq M$, which implies that $M+I=R$ because $M$ is maximal. Thus there exists $i\in I$ and $m_{1}\in M$ such that $1=m_{1}+i$. Likewise we see that $J+M=R$, so there exists $j\in J$ and $m_{2}\in M$ such that $1=j+m_{2}$. Now, \begin{align} 1&=1\cdot 1\\ &=(m_{1}+i)(j+m_{2})\\ &=m_{1}j+m_{1}m_{2}+ij+im_{2} \end{align} but $m_{1}j,m_{1}m_{2},im_{2}\in M$ because $M$ is a two-sided ideal. Also $ij\in M$ since $ij\in IJ$ and $IJ\subseteq M$. But this implies that the sum, and therefore $1\in M$, which contradicts the maximality of $M$. Therefore we conclude that $M$ is a prime ideal.