First, let me explicitly assume $R$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$.
Proposition. The Krull dimension of $R$ is at most $1$.
Proof. Let $\mathfrak{m} = (t)$, and let $\mathfrak{p}$ be prime. We know $\mathfrak{p} \subseteq \mathfrak{m}$, so it is enough to show that either $\mathfrak{p} = \mathfrak{m}$ or $\mathfrak{p} = (0)$. Suppose $\mathfrak{p} \ne \mathfrak{m}$: then $t \notin \mathfrak{p}$. Let $a_0 \in \mathfrak{p}$. For each $a_n$, because $\mathfrak{p}$ is prime, there exists $a_{n+1}$ in $\mathfrak{p}$ such that $a_n = a_{n+1} t$. By the axiom of dependent choice, this yields an infinite ascending sequence of principal ideals:
$$(a_0) \subseteq (a_1) \subseteq (a_2) \subseteq \cdots$$
Since $R$ is noetherian, for $n \gg 0$, we must have $(a_n) = (a_{n+1}) = (a_{n+2}) = \cdots$. Suppose, for a contradiction, that $a_0 \ne 0$. Then, $a_n \ne 0$ and $a_{n+1} \ne 0$, and there is $u \ne 0$ such that $a_{n+1} = a_n u$. But then $a_n = a_{n+1} t = a_n u t$, so cancelling $a_n$ (which we can do because $R$ is an integral domain), we get $1 = u t$, i.e. $t$ is a unit. But then $\mathfrak{m} = R$ – a contradiction. So $a_n = 0$. $\qquad \blacksquare$
Here's an elementary proof which shows why we can reduce to the case where $R$ is an integral domain.
Proposition. Any non-trivial ring $A$ has a minimal prime.
Proof. By Krull's theorem, $A$ has a maximal ideal, which is prime. Let $\Sigma$ be the set of all prime ideals of $A$, partially ordered by inclusion. The intersection of a decereasing chain of prime ideals is a prime ideal, so by Zorn's lemma, $\Sigma$ has a minimal element. $\qquad \blacksquare$
Thus, we can always assume that a maximal chain of prime ideals starts at a minimal prime and ends at a maximal ideal. But if $R$ is a noetherian local ring with principal maximal ideal $\mathfrak{m}$ and $\mathfrak{p}$ is a minimal prime of $R$, then $R / \mathfrak{p}$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$, and $\dim R = \sup_\mathfrak{p} \dim R / \mathfrak{p}$, as $\mathfrak{p}$ varies over the minimal primes.
Update. Georges Elencwajg pointed out in a comment that the first proof actually works without the assumption that $R$ is a domain, because $(1 - u t)$ is always invertible.
Let $I$ be an ideal of $R$, $I\ne 0,R$. Then $I\subseteq\mathfrak m$, but $I$ can't be contained in all the powers of $\mathfrak m$, otherwise $I=0$, contradiction. So there is a greatest $i\ge 1$ such that $I\subseteq \mathfrak m^i$, and $I\not\subseteq\mathfrak m^{i+1}$. Let's show that $I=\mathfrak m^i$. Since $I\not\subseteq\mathfrak m^{i+1}$ there is $a\in I\setminus\mathfrak m^{i+1}$. If $\mathfrak m=(\pi)$, then $a=\pi^ia'$. Since $a\notin\mathfrak m^{i+1}$ we get $a'\notin\mathfrak m$, so $a'$ is invertible. From $a=\pi^ia'$ we get now that $\pi^i\in I$ and therefore $I=\mathfrak m^i$.
Best Answer
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 $I = (α^n) = m^n$.