Let $α$ be the generator of $m$, i.e. $m = (α)$.
Hint: First try to prove for any nonzero nonunit $x ∈ R$ that $(x) = (α^n)$ for some $n ∈ ℕ$, then use the fact that any ideal $I ≠ 0$ is finitely generated to conclude what you want to show.
I’ve already done that, but now realized you maybe wanted to do this yourself. But I will leave below what I already did, in case you want to take a peek.
Let $x ∈ R$ be nonzero nonunit, i.e. $(x) ≠ R$ and $(x) ≠ 0$. Then there’s a maximal $n ∈ ℕ$ such that $x ∈ m^n$ (because $x$ has to lie in the only maximal ideal $m$ at least and $\bigcap_{n ∈ ℕ} m^n = 0$). Write $x = rα^n$. Now $r$ cannot be in $m$, or else $n$ wouldn’t be maximal. Therefore $r ∈ R\setminus m = R^×$ and so $(x) = (α^n)$.
Since $R$ is noetherian, you can write any nontrivial ideal $I ≠ 0$ using nonzero generators $x_1, …, x_s ∈ R$ as $I = (x_1,…,x_s)$. Do the above argument for the generators: $I = (α^{n_1}, …, α^{n_s})$. Take $n = \min \{n_1, …, n_s\}$, then $I = (α^n) = m^n$.
You need that $R$ is noetherian, else there are counterexamples.
E.g., take $R = K[x_i]_{i \in \mathbb{N}}/(x_i | i \in \mathbb{N})^2$.
If $R$ is Noetherian, this is a special case of Lemma 10.59.4 here which states that a Noetherian ring of dimension $0$ is Artinian.
Best Answer
Nakayama's Lemma holds for arbitrary modules (without f.g. assumption): $M/\mathfrak{m}M=0$ implies $M=0$. Of course this is purely formal and holds for every nilpotent ideal.