$a$ is a non-zerodivisor, in particular $a$ is not contained in a minimal prime of $R$. In particular any prime over $(a)$ is of height at least $1$, hence a maximal ideal, since $R$ has dimension $1$.

Proof that a minimal prime consists of zero divisors:

Let $\mathfrak p_1, \dotsc \mathfrak p_s$ be the minimal primes of a noetherian ring $R$. $s$ is finite due to the noetherian property.
We know the nilradical of $R$ is the intersection of the minimal primes, hence contains the product of the minimal primes. So $\mathfrak p_1 \dotsb \mathfrak p_s$ is nilpotent, which means there are minimal integers $n_1, \dotsc, n_s$, such that
$$\mathfrak p_1^{n_1} \dotsb \mathfrak p_s^{n_s}=0.$$

Given any $x \in \mathfrak p_1$, choose $0 \neq y \in \mathfrak p_1^{n_1-1} \dotsb \mathfrak p_s^{n_s}$ and we have the desired $xy=0$.

Of course the argument requires $n_i \geq 1$ for all $i$. But this is an easy exercise. Literally one has to show that $\mathfrak p_2 \dotsb \mathfrak p_s$ (leaving out one of the minimal primes) is not nilpotent.

After some thought, one does not need the noetherian property: Let $\mathfrak p$ be a minimal prime. $\mathfrak p R_{\mathfrak p}$ is the only prime ideal of $R_{\mathfrak p}$, hence it coincides with the nilradical (This does not mean that $\mathfrak p R_{\mathfrak p}$ is nilpotent in the nonnoetherian case, but any element is nilpotent). Let $x \in \mathfrak p$. Then by the previous observation $x^n=0$ in $R_{\mathfrak p}$. The definition of localization yields some $t \notin \mathfrak p$, such that $x(x^{n-1}t)=x^nt=0$. If $n$ is chosen minimal we have $0 \neq x^{n-1}t$, hence $x$ is a zerodivisor.