There exists no ideal $I$ such that $M^2 \subset I\subset M$ for $M$ a principal maximal ideal I am looking for an answer to the question

Let $R$ be a commutative ring with $1$, and let $M = (m)$ be a principal maximal ideal. Prove that there is no ideal $I$ such that $M^2 \subset I \subset M$. 

I was given the following proof using module theory:

The quotient $\overline{M} = M/M^2$ is a module over the field $K = R/M$. Moreover, it is one-dimensional because $M$ is principal. Therefore the only submodules are $0$ and $\overline{M}$. Lifting back to $R$, this shows that the only ideals contained in $M$ and containing $M^2$ are $M$ and $M^2$. 

Is there a proof of this question without appealing to module theory? 
I understand that $M^2 = (m^2)$, and it "makes sense" that $R/M^2$ is a local ring with its unique maximal ideal as $(m + M^2)$ since all elements have the form $a + bm + M^2$, but I'm having difficulty proving this. 
 A: The proof “with elements” is the same as the proof “with vector spaces”: it's just a question of translation.
The module $M/M^2$ is a one dimensional vector space over $R/M$. 
Translation. $M=(m)$.
If $I$ is an ideal between $M^2$ and $M$, $I\ne M^2$, then $I/M^2$ is a nonzero vector subspace of the one-dimensional space $M/M^2$, so it is the same as $M/M^2$. 
Translation. Let $x\in I$, $x\notin M^2$. Then $x=rm$, for some $r\in R$, because $M=(m)$; since $x\notin M^2$, we have $rm\notin M^2$, in particular $r\notin M$; then $r+M$ is invertible in $R/M$, which means there are $s\in R$ and $y\in M$ with $1=rs+y$. Thus
$$
m=1m=(rs+y)m=s(rm)+ym\in I+M^2=I
$$
and therefore $I=M$.
Using vector spaces is just easier.
A: Suppose $I$ contained $rm$, with $r\not\in M$. Since $M$ is maximal, we can write $1=rt+sm$ for $r,s\in R$. If we further assume $m^2\in I$, then we have $m=(rm)t+s(m^2)\in I$, and so $M\leq I$.
Thus if $M^2\le I\le M$, either $I=M$, or no such element $rm$ exists in $I$, and $I=M^2$.
