Ring is Noetherian if it admits a faithful finitely generated module with ACC on submodules generated by ideals 
I have a (commutative unitary) ring $A$ and a faithful finitely generated module $M$ over $A$ which satisfies ACC on submodules of the form $IM$, where $I\subset A$ is an ideal. I want to prove that $A$ is Noetherian.

I have already proved that, because $M$ is finitely generated, if $M$ is a Noetherian $A$-module, then $A/Ann(M)$ is a Noetherian ring. So, given my hypothesis, to prove that $A$ is Noetherian as a ring it suffices to prove that $M$ is Noetherian as $A$-module.
I've been given a sketch of proof, but I'm stuck with filling in the details. Here's the sketch:
Proceed by contradiction and assume $M$ is not Noetherian. Consider the set $\{IM \mid I\subset A \text{ is an ideal and }M/IM \text{ is not Noetherian}\}$ and reduce to the case where $M$ is not Noetherian, but for every ideal $I\neq 0$, $M/IM$ is Noetherian. Then consider the set $\Gamma = \{N\subset M \mid M/N \text{ is a faithful } A\text{-module}\}$, find a maximal element in $\Gamma$ and conclude.
Here's my attempt at using the hints I've been given: using ACC on submodules of the form $IM$ I found a maximal submodule $JM$ such that $M/JM$  is not Noetherian, so I thought substituting $M$ with $M/JM$, but $M/JM$ is clearly not faithful so I don't think that's the right approach.
Then I tried to assume the reduction to just $M$ not Noetherian had been made and tackled the set $\Gamma$. I managed to find a maximal element (using Zorn's lemma), but I'm clueless on how to use that maximal element to conclude.
I don't need the solution straight away, just some help to understand the hints better would be very much appreciated.
(I hope I explained everything clearly cause english is not my first language.)
 A: Edit (12/31): This result turns out to have a bit of history. It was published in 1973, though there were other formulations around prior to that [1]. One finds that this result has been reproduced in Matsumura's textbook, Commutative Ring Theory [2]. The method of proof is essentially the same as what I had answered prior to looking it up, with the first step proceeding in basically the same way. The second step is done a little differently, so I provide both in this answer. Since the approaches are similar in content, I made the answer community wiki.
First step: Let $IM$ be maximal such that $M/IM$ is not Noetherian as an $A$-module. Let $I_0=\operatorname{Ann}(M/IM)$. Then, $I_0M=IM$ and $M/I_0M$ is a faithful $A/I_0$-module. Thus, replacing $M$ with $M/I_0M$ and $A$ with $A/I_0$, we may assume $M/JM$ is Noetherian for all nonzero ideals $J$ of $A$, but $M$ is not Noetherian. We will arrive at a contradiction by showing that this $A$ is Noetherian (and thus this $M$ is Noetherian).
Second step: (The approach in the references) By Zorn's Lemma, we get a maximal submodule $N\subset M$ such that $M/N$ is a faithful $A$-module. We may assume $M/N$ is not Noetherian. So, replacing $M$ with $M/N$, we have
(a) $M$ is not Noetherian
(b) If $I$ is a nonzero ideal of $A$, then $M/IM$ is Noetherian
(c) $M$ is faithful, and for all nonzero submodules $N\subset M$, the $A$-module $M/N$ is not faithful.
Let $N$ be any submodule of $M$. We will prove $N$ is finitely generated. If $N=0$, then that is the case. Suppose $N\ne 0$. By statement (c), there exists $0\ne a\in A$ such that $aM\subset N$. Because $N/aM$ is a submodule of $M/aM$, which is Noetherian, $N/aM$ is finitely generated. Furthermore, if $x_1,\dots,x_n$ generate $M$, then $ax_1,\dots,ax_n$ generate $aM$. So, $aM$ is finitely generated. Therefore, $N$ is finitely generated. It follows that $M$ is Noetherian, violating statement (a). This is the desired contradiction.
(A different approach) Any ascending chain of submodules $M_i$ of $M$ such that $M_i\neq M_{i+1}$ for all $i$ has the property that $M/M_i$ is faithful for all $i$ (otherwise, $M_i\supset IM$ eventually and we know $M/IM$ is Noetherian, contradiction). Thus, if $N$ is a maximal submodule of $M$ such that $M/N$ is a faithful $A$-module, then $M/N$ is Noetherian. Hence, $A$ is Noetherian. This is a contradiction.
Having arrived at a contradiction, we conclude that our original $M$ is Noetherian, as wanted.
References:
[1] E. Formanek. "Faithful Noetherian Modules". Proc. AMS. 1973. 41(2): 381-3.
[2] H. Matsumura. Commutative Ring Theory. 2006. p. 18 (Theorem 3.6).
