In my book, it is claimed without proof that if a ring (commutative with identity) $R$ has zero divisors, then $R[x]$ has invertible polynomials of degree greater than $0$. They give the following example in $\mathbb{Z}_4$: $$(1+2x)(1+2x)=1$$

This seems like a pretty lucky example: in this ring we have a zero divisor whose sum with itself is $0$. In general this doesn't seem like an easy question to tackle. First i thought that maybe could find some $a$ such that $a^2 = 0$ and use the factorization $(1+ax)(1-ax)$ but in general such $a$ does not exist: $\mathbb{Z}_6$ is a counterexample. No polynomial of form $(1+ax)$ has a degree $1$ inverse in $\mathbb{Z}_6[x]$ either. Trying out higher degrees seems like a nightmare.