Ring of formal power series over a principal ideal domain is a unique factorisation domain

Another answer to the problem given by the OP:

If $R$ is a PID, then $R[[X]]$ is a UFD.

The key to prove the above result is to use a theorem known as Kaplansky's criterion for UFDs:

Theorem: An integral domain $D$ is a UFD if only if every nonzero prime ideal of $D$ contains a prime element.

I'm not going to write all the details because I don't want to write a long post. In order to use Kaplansky's criterion we need the following lemma:

Lemma: Let $A$ be a commutative unitary ring. Let $\phi\colon A[[X]]\rightarrow A$ be the map given by $$f=\sum_{i=0}^{\infty}a_iX^i\mapsto a_0.$$ Then $\phi$ is a surjective ring homomorphism. Moreover if $\mathfrak{P}$ is a prime ideal of $A[[X]]$, then $\mathfrak{P}$ is finitely generated in $A[[X]]$ if only if $\phi(\mathfrak{P})$ is finitely generated in $A$.

A proof of the above lemma can be found in Sharp's book "Steps in Commutative Algebra" or in Watkins' book "Topics in Commutative Ring Theory". In such proofs it's shown the following:

If $X\in \mathfrak{P}$, then $\mathfrak{P}$ is generated by the same elements that generate $\phi(\mathfrak{P})$ plus $X$; and if $X\notin \mathfrak{P}$, then the number of generators of $\mathfrak{P}$ is the same as that of $\phi(\mathfrak{P})$.

Finally, to prove the theorem given at the beginning of the post we proceed as follows. We first note that $X$ is a prime element of $R[[X]]$. Next, given a nonzero prime ideal $\mathfrak{P}$, if $X\in \mathfrak{P}$, then $\mathfrak{P}$ contains the prime element $X$. Otherwise, by the lemma $\mathfrak{P}$ is f.g. with the same number of generators as the ideal $\phi(\mathfrak{P})$, but $R$ being a PID implies that $\phi(\mathfrak{P})$ is principal, so $\mathfrak{P}$ is principal too, i.e., there is some $f\in R[[X]]$ such that $\mathfrak{P}=(f)$. Since $\mathfrak{P}$ is a prime ideal is easy to see that $f$ is a prime element, so $\mathfrak{P}$ contains the prime $f$. Hence, by Kaplansky's criterion $R[[X]]$ is a UFD.

As a side remark, the lemma that we've used is also applied (together with Cohen's theorem) to prove the analogue of Hilbert basis' theorem for formal power series.

The set of irreducible elements of $\mathbb{Z}[[x]]$ is not known, and is more complicated as it seems.

You might find some interesting families of irreducible elements in this paper :

http://www.ijma.info/index.php/ijma/article/view/2357

However, it does not answer the original question.