 Solution

Let $f\in R$ be irreducible. If $\mathrm{deg}f=0$, then $f\in \mathbb{Z}$. A nonzero integer that is neither a unit nor, up to sign, a prime number factors nontrivially in $R$, so necessarily $f=\pm p$ for a prime $p$.

Now suppose $\mathrm{deg}f\ge 1$ and write $m=f(0)\in \mathbb{Z}$. If $m=0$, thenand both factors lie in $R$ and are nonunits. If $|m|>1$, choose a prime $p\mid m$; then $f=p(f/p)$ is again a factorization into two nonunits of $R$, since $(f/p)(0)=m/p\in \mathbb{Z}$. Both cases contradict irreducibility. Hence