Let $R$ be a commutative Noetherian ring with unique maximal ideal $\mathfrak{m}$. Let $N$ be a finitely generated $R$-module with the property that every exact sequence $0 \to L \to M \to N \to 0$ remains exact tensoring with any $R$-module. Prove that the $R$-module $N$ is free.
Here is my thoughts of this problem: We know that every submodule of a finitely generated $R$-module is finitely generated, since $R$ is Noetherian. We also know that $R/\mathfrak{m}$ is a field since $\mathfrak{m}$ is maximal, and thus tensoring the exact sequence with this field and using the fact that $R/\mathfrak{m} \otimes_R L \cong L/\mathfrak{m}L$ and similarly for $M$ and $N$ gives us the exact sequence $0 \to L/\mathfrak{m}L \to M/\mathfrak{m}M \to N/\mathfrak{m}N \to 0$.
Here are my further thoughts. Since $N$ is finitely generated, there exists some $n \in \mathbb N$ and a short exact sequence $R^n \to N \to 0$ (i.e. a surjection $\varphi$ from $R^n$ to $N$). This sequence can be extended to the short exact sequence $0 \to \ker \varphi \to R^n \to N \to 0$ so that tensoring with $R/\mathfrak{m}$, we get the short sxact sequence $0 \to \ker \varphi/\mathfrak{m} \ker \varphi \to (R/\mathfrak{m})^n \to N/\mathfrak{m}N \to 0$. If $\mathfrak{m}N = N$, then by Nakayama's lemma we have that there exists $m \in \mathfrak{m}$ such that $(1-m)N = 0$. If $1-m \not\in R^\times$, then since $\mathfrak{m}$ is the unique maximal ideal of $R$, we may get $\langle 1-m, \mathfrak{m} \rangle$ is an ideal of $R$ containing $\mathfrak{m}$ and thus $1-m \in \mathfrak{m}$, so that $1 = (1-m) + m \in \mathfrak{m}$, a contradition. Thus $1 - m$ is invertible, and thus $N = 0$, which is free. Thus, we may assume that $\mathfrak{m}N < N$. I am not sure where to go from here.