Let $M$ be an $R$-module with a minimal generating set $S$, where $|S|\geq 2$. Then $M$ has at least two maximal submodule. I saw the following result somewhere with its proof:
Result: Let $M$ be an $R$-module with a minimal generating set $S$, where $|S|\geq 2$. Then $M$ has at least two maximal submodules.
Proof: Let $x_1, x_2$ be two distinct element in $S$. Since $S$ is a minimal generating of $M$ we infer that $x_1R\nsubseteq x_2R$ and $x_2R\nsubseteq x_1R$. Hence by Zorn's Lemma there are two distinct maximal submodules, $N, P$ say, with $x_1\in N, x_2\notin N$ and $x_2\in P, x_1\notin P$.
I have doubt in the last line of the proof. My doubt: why $N$ and $P$ are maximal submodules of $M$? They are just maximal member of a class of submodules. Please explain it.
 A: The result is true even if the minimal generating set $S$ of $M$ is infinite.
Suppose $x\in S$. Then the submodule of $M$ generated by $S\setminus\{x\}$ does not contain $x$ (as otherwise $S$ is not a minimal generating set). The union of a chain of submodules that do not contain $x$ does not contain $x$. So by Zorn's Lemma there is a submodule $N$ that contains $S\setminus\{x\}$, maximal subject to not containing $x$. Any submodule properly containing $N$ must contain $S$ and so must be equal to $M$. So $N$ is a maximal submodule of $M$.
So if $x_1$ and $x_2$ are distinct elements of $S$ then there is a maximal submodule containing $S\setminus\{x_1\}$ but not $x_1$, and another maximal submodule containing $S\setminus\{x_2\}$ but not $x_2$.
A: I’m writing this under the impression that the minimal generating set is finite, because otherwise it’s not clear to me why any maximal submodules should exist.
You're right, the application of Zorn's lemma is inadequately justified.  It is not even clear what it is being applied to.  It's also unclear that just arguing with $x_1$ and $x_2$ is enough. It seems like you have to involve the (potentially very large collection of) $x_i$'s other than those two.

What's releveant is that by Zorn's lemma, any finitely generated module has a maximal submodule.
Now, to make that useful, let $x_1,\ldots x_n$ be a minimal generating set for $M$.  Now, consider the submodule generated by $\{x_2,\ldots x_n\}$. Since this is less than a minimal generating set, it must be contained in a maximal submodule $N_1$ of $M$, and $x_1$ is certainly not in $N_1$ (or else all the $x_i$'s together could not generate $M$.)
For the same reason, there's a maximal submodule $N_2$ of $M$ containing $\{x_i\mid i\neq 2\}$, and $x_2\notin N_2$.
Now it's clear that you have two distinct maximal submodules.
By this reasoning, you have at least $n$ distinct maximal submodules.
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As the other solution now points out, using the trick of finding a maximal submodule leaving out one generator does always yield a maximal submodule, so it does work when $S$ is infinite.  That means (counterintuitively, to me) that all examples of modules with no maximal submodules do not have minimal generating sets.  I let that disbelief cloud what should have been a pretty straightforward check.
I’m leaving the answer up because it contains the feedback on the users original attempt.
