This depends heavily on what your definition of "Noetherian" is. Are you working with the ascending chain condition on ideals?
If you start with an infinite nondecreasing chain $A_i\subseteq A_{i+1}$ of proper ideals (in a ring with identity), then $\bigcup_{i=1}^\infty A_i=(a)$ is a proper ideal containing all the $A_i$'s.
But $a\in A_j$ for some $j$, and that would mean that $(a)\subseteq A_j$. In that case, $A_j=\bigcup A_i=A_k$ for every $k\geq j$.
This says that all such chains stabilize, that is, the ring satisfies the ascending chain condition on ideals.
The same reasoning shows you that if you just suppose all ideals are finitely generated, then the ascending chain condition holds.
Are you using the maximum condition on ideals?
I think, perhaps, it was intended for you to use Zorn's Lemma to prove that ACC on ideals implies this. For if you are given a nonempty set of ideals of $R$, Zorn's lemma and the ACC together imply that every chain in the set is bounded, and hence the whole set has a maximal element.
The converse implication (maximal condition implies ACC) always holds, of course: given any chain, the maximal condition implies the chain has a greatest element, and that would make the chain stable.
Despite the name, Zorn's Lemma is just the Axiom of Choice in another guise. You are free to assume it holds, or not, and we often assume it. Using it is easy: if its hypotheses are satisfied, then its conclusion holds. To do this, you need to validate that ascending chains in a poset are bounded in that set, and then you magically get a maximal element somewhere in the poset.
Let $g\in G$. Note that because $M$ is abelian, $\langle C_M(g),g\rangle$ is abelian. If $C_M(g)=M$, then $M\leq \langle C_M(g),g\rangle$, so the maximality of $M$ gives $M=\langle C_M(g),g\rangle$, hence $g\in M$.
By contrapositive, if $g\notin M$, then $C_M(g)\neq M$.
Best Answer
Zorn's lemma is not needed.
I'll use additive notation for the abelian group $G$.
Indeed, let $(H_\alpha)$ be a family of divisible subgroups and $H=\sum_\alpha H_\alpha$. Let $x\in H$; then $x=\sum_{i=1}^n x_{\alpha_i}$, for $x_{\alpha_i}\in H_{\alpha_i}$. If $m>0$ is an integer, then for each $i$, there is $y_i\in H_{\alpha_i}$ with $$ my_i=x_{\alpha_i} $$ Set $y=\sum_{i=1}^ny_i$; then $y\in H$ and $my=x$.
Thus you can consider the sum $D$ of all divisible subgroups of $G$, which is a divisible subgroup and the larges such.
Let's prove that $G/D$ has no divisible subgroup except $\{0\}=D/D$.
To this end we show that if $A$ is a divisible subgroup of $B$ and $B/A$ is divisible, then also $B$ is divisible. Let $x\in B$ and $m>0$. By assumption there is $y\in B$ such that $$ x+A=my+A $$ (since $B/A$ is divisible). This means that $x-my\in A$, which is divisible as well, so there is $z\in A$ with $mz=x-my$. Hence $x=m(y+z)$ as desired.