Hint: Let $d$ be the Krull dimension of $A$. Show that $x$ is part of a set $x,x_2,\ldots,x_d\in A$ such that $(x,x_2,\ldots,x_d) = m$ iff $x\in m-m^2$. From this (and using the fact that there is a bijection between the maximal ideals of $A$ containing $(x)$ and the maximal ideals of $A/(x)$), you can see that $A/(x)$ is a local ring with maximal ideal $(x_2,\ldots,x_d)$, which has Krull dimension at most $d-1$ by Krull's Principle Ideal Theorem. The dimension is in fact exactly $d-1$, as if we had fewer than $d-1$ generators for $m/(x)$ then adjoining $x$ gives you fewer than $d$ generators for $m$, violating Krull's PIT. Since $m/(x)$ is generated by $d-1$ elements, this makes $A/(x)$.