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.
I suggest you R. Y. Sharp's book: Steps in Commutative Algebra.
By Theorem 15.4 (Krull's generalized principal ideal theorem), $\dim R$ is less than or equal to "the number of elements in each minimal generating set for $m$", which is finite since $R$ is Noetherian. And by Proposition 9.3 this is equal to $\dim_k m/m^2$.
Best Answer
If $A$ a Noetherian ring of Krull dimension $0$, it implies that $A$ an Artinian ring and that every prime ideal is maximal.
So you only need to check that $A$ has finitely many maximal ideal:
Consider the set $\Sigma$ of all finite intersections of maximal ideals. $\Sigma$ is non-empty and thus has a minimal element $\mathfrak m_0=\bigcap_{i=0}^n\mathfrak m_i$ (because $A$ is Artinian).
Let $\mathfrak{m}$ be a maximal ideal of $A$, then $\mathfrak m \cap \mathfrak m_0\subset \mathfrak m_0$ and these elements are in $\Sigma$. By minimality of $\mathfrak{m}_0$, you get $$\mathfrak m\cap \mathfrak m_0 = \mathfrak m_0.$$ Hence, $\mathfrak m$ must contain one of the $\mathfrak m_i$ and since these ideals are maximal, it must be equal to one of the $\mathfrak m_i$.