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
Prologue
For any commutative ring $R$ and any ideal $I\subsetneq R$ we have $$ \dim(R/I)+ht(I)\leq \dim (R) \quad (*)$$ This does not assume $R$ noetherian, nor local, nor... but just follows from the definitions.
Inequality
Suppose now that $(A,\mathfrak m)$ is local noetherian.
The trick is to use that $\dim(A)$ is the smallest number of elements in $\mathfrak m$ generating an $\mathfrak m$-primary ideal (cf. Atiyah-Macdonald Theorem 11.14). Let's do that for $A/xA$:
If $\bar x_1,...,\bar x_k\in \mathfrak m/xA$ generate an $\mathfrak m/xA$-primary ideal, then $x, x_1,..., x_k$ generate an $\mathfrak m$-primary ideal and this immediately yields the required inequality $$\dim(A)\leq \dim(A/xA)+1 \quad (**)$$
Equality
The Prologue implies that equality in $(**)$ will hold if $ht(xA)=1$.
The principal ideal theorem says that we always have $ht(xA)\leq 1$.
Now to say that $ht(xA)=0$ means that $x\in \mathfrak p$ for some minimal ideal $\mathfrak p$.
But it is well known that minimal ideals consist of zero divisors (= non-regular elements).
Hence if $x$ is regular we have $ht(xA)=1$ (since we don't have $ht(xA)=0$ !) and the required equality follows $$ \dim(A)= \dim(A/xA)+1 \quad (***)$$