Suppose $A$ is a commutative ring with $1\neq0$ satisfying the property that $A_\mathbf{m}$ has no nonzero nilpotent elements for any maximal ideal $\mathbf{m}$, where $A_\mathbf{m}=S^{-1}A\quad \text{ and }\quad S=A-\mathbf{m}.$ Prove that $A$ has no nonzero nilpotent elements.
My attempt at the solution was to prove the contrapositive, i.e., suppose $A$ has a nonzero nilpotent element $x$. Then I would like to show that $A_\mathbf{m}$ has a nonzero nilpotent element for some maximal ideal. When I'm using the definition for the equivalence class of fractions, i.e., $\frac{a}{b}=\frac{a'}{b'}\quad \text{if and only if}\quad (ab'-a'b)u=0$ for some $u\in S$, I'm having trouble showing that $\frac{x}{s}\neq\frac{0}{1}$. This is because the definition reduces us to $x\cdot u=0$, and it very well seems that $u$ could be the element that makes $x$ a zero divisor. Is there some way to construct an ideal that contains no zero divisors of $x$? My original thought was to take $\mathbf{m}$ to be the maximal ideal containing all of the zero divisors of $x$, but the smallest ideal containing all of the zero divisors very well could be the whole ring (as in the case of $\mathbb{Z}/15\mathbb{Z}$).
Any help or ideas regarding the proof would be very helpful!