I wonder if my proof of the above statement is correct. I did not find my proof in the linked answers, so now I am not sure if it is correct because it seems to be a little bit too easy.
Let be $f \colon M \to M$ a surjective endomorphisms of an $R$-module $M$. Then $M$ is an $R[x]$-module by letting $x$ act as $f$. Consider the ideal $I=(x)$. Then $IM=M$ holds since $f$ is surjective and $x \in I$. We can apply Nakayama's lemma because $M$ is a finitely generated $R[x]$-module. It follows that there exists an element $g \equiv 1 \ (\mathrm{mod}\ I)$ such that $gM=0$. Because $g \equiv 1 \ (\mathrm{mod}\ I)$ means that $g$ has constant term equal to $1$, it follows $n = 0$ for every $n \in \ker f$.
Surjective endomorphisms of finitely generated modules are isomorphisms
Surjective endomorphisms of finitely generated modules are injective
Best Answer
You can write $g$ as $1+h\cdot x$ for some $h \in R[x].$ If $m\in M$ belongs to the kernel of $f,$ note that $0=(1+h\cdot x)m=m+h\cdot f(m)=m.$ Hence, $f$ is an injection, so an isomorphism. This is the way I have usually seen the proof finished. You can see it as Theorem 2.4 in Matsumura's Commutative Ring Theory. You might want to check Vasconcelos' original paper, "On finitely generated flat modules," Trans. Amer. Math. Soc. 138 (1969) 505-512.