I've been trying to solve the above problem. There seems to already be some work regarding zero divisors in polynomial rings over here, but I'm not sure it is applicable to me, because it looks specifically for some $r \in R$ such that for a zero divisor $F \in R[x]$ holds $F \cdot r = 0$.
I, on the other hand, am looking to prove that if $R[x]$ has zero divisors, it implies $R$ has zero divisors.
My ideas so far
I've thought of the following: Let $p(x), q(x) \in R[x], \quad p(x) \neq 0 \neq q(x), \quad p(x) \cdot q(x) = 0$.
Since the product of $p(x)$ and $q(x)$ is zero, it shouldn't matter which $x \in R$ you choose. The product should still be zero regardless. So, freely choosing a single $x \in R$, we should be able to reduce both $p(x)$ and $q(x)$ to a value in $R$, with the equation still holding. As such, we'd have found our zero divisors in $R$.
Unfortunately, we have to watch out so that $x$ isn't a zero of either $p(x)$ or $q(x)$, or else we'd reduce one of them to $0$ and as such we wouldn't be finding a zero divisor. This leaves me stuck.
Obviously, for my approach I'd have to show that there is an $x \in R$, so that neither $p(x)$ nor $q(x)$ reduce to $0$.