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 $I = (α^n) = m^n$.
Best Answer
Here is an elementary proof (which is of course much easier than the proof of the deeper theorem of Kaplansky), which does not invoke any theorems in commutative algebra besides Nakayama:
Let $I \subset (m)$ be a non-zero ideal. Define $n := \min \{ s \geq 1 | \exists x \in I \text{ such that } x \in (m)^s \setminus (m)^{s+1} \}$.
First, we have to show that $n < \infty$, i.e. we have to show that $I$ is not contained in $\bigcap\limits_{s \geq 1} (m)^s$, i.e. we have to show $\bigcap\limits_{s \geq 1} (m)^s=0$. This is clear with Krull's intersection theorem, but I don't want to invoke any theorem, so I will give an argument in this case:
Let $a$ be contained in that intersection, in particular $a=mb$ for some $b \in A$. I claim that $b$ is also contained in that intersection: If not, $b$ is non-zero in some $(m)^s/(m)^{s+1}$. Since this is a one-dimensional vector-space, we obtain that $b$ generates $(m)^s/(m)^{s+1}$. By Nakayama, $b$ generates $(m)^s$, i.e. $a$ generates $(m)^{s+1}$. In particular $a \notin (m)^{s+2}$, contradiction!
Thus, we have shown $(m) \bigcap\limits_{s \geq 1} (m)^s = \bigcap\limits_{s \geq 1} (m)^s$, i.e. $\bigcap\limits_{s \geq 1} (m)^s=0$ by Nakayama. This is where we need the noetherian hypothesis, because we need to guarantee that $\bigcap\limits_{s \geq 1} (m)^s$ is a priori finitely generated to invoke Nakayama.
Now it is very easy to show that $I=(m)^n=(m^n)$ holds: By the minimality of $n$, we have $I \subset (m)^n$ and we have some $x \in I$ with $x \notin (m)^{n+1}$. Again invoking Nakayama, we get $(x)=(m)^n$, i.e. $I \subset (m)^n=(x) \subset I$. The proof ends here.