= Solution
Let $f\in R$ be irreducible. If $\deg f=0$, 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 $\deg f\geq1$ and write $m=f(0)\in\mathbb Z$. If $m=0$, then
$$
f=2(f/2),
$$
and 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
$$
\boxed{\deg f\geq1\quad\Longrightarrow\quad f(0)=\pm1.}
$$
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