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.
Best Answer
The statement is wrong. Consider for example the integers, then $\dim(\mathbb{Z})=1$ and it is noetherian as it is a PID but for any prime number we have that $\mathbb{Z}/p\mathbb{Z}$ is a field, thus has dimension zero with the only prime ideal $(0)$. Your problem here is that when factoring out a nonzero prime ideal your dimension reduces, in this case by one as any nonzero prime ideal in a domain of dimension one has height one. So you always end up with a field where the maximal ideal is just $(0)$.