The observation in this answer and this comment suggest the following:
If $R$ is a domain and $f \in R[x]$ is monic and not a unit in $R[[x]]$, then if $f$ is irreducible in $R[x]$ it is also irreducible in $R[[x]]$.
Is this true, and if so how do we prove it? It doesn't seem obvious.
The statement when $R$ is a field seems trivial, since if $f$ is not a unit as a power series it is always reducible, since it has a zero constant term and so it can be factored as some power of $x$ times another polynomial.
Best Answer
It is false, e.g. this answer shows the linear polynomial $\,ab+x\,$ is reducible in $R[[x]]$ if $\,a,b\neq 0\,$ are comaximal nonunits, giving an explicit factorization via a Catalan series $f(x)\in\Bbb Z[[x]],\,$ e.g.
$$\begin{eqnarray}{} 6+ x\, &=&\ (3 - x\:\!f(x))\ \ (2 + x\:\!f(x)) \\[0.5em] &=&\ \dfrac{5\!-\! \sqrt{1\!-\!4\:\!x}}{2}\,\ \dfrac{5\! +\! \sqrt{1\!-\!4\:\!x}}{2}\\ \end{eqnarray}\qquad\qquad$$
It is not difficult to construct algorithms for irreducibility testing and factorization in the UFD $R[[x]]$ over a PID $R,\,$ e.g. see J. Elliot: Factoring formal power series over principal ideal domains.