Denote by $|\cdot |$ the $p$-adic norm, and let $$\mathbb Z_p \{z\}=\left\{\sum_{n=0}^\infty a_nz^n;\ a_n\in \mathbb Z_p ,\ |a_n|\underset {n\rightarrow \infty} {\longrightarrow 0} \right\}$$ the ring of converging power series over the p-adic integers with indeterminate $z$. I need to show that this ring is a unique factorization domain.
What I know is that $\ \mathbb Q_p\{z\}$ is a principal ideal domain, and in fact every element $\ f\in\mathbb Q_p\{z\}$ has a unique representation $f=qu$ where $\ q\in\mathbb Q_p[z],\ u\in \mathbb Q_p\{z\}^\times$. Another thing is that for archimedean valuations in general this isn't true. For example, the ring of converging power series over $\ \mathbb C$ with the usual absolute value is not a UFD, as $\sin z$ which is holomorphic in the unit disc don't have a unique representation as finite product of irreducibles (the Laurent series of $\sin z$ around the origin is an infinite product of irreducibles).
I think this is supposed to be true for $\mathbb Z_p\{z\}$, and the fact that $\mathbb Q_p\{z\}$ is a PID is crucial for the proof, but I haven't been able to see how.
I can only give an inefficient answer; others will do much better.
I hope you know about and are thinking of Newton Polygon. Remember that it’s described by using the additive valuation on $\Bbb Q_p$, $v(p)=1$, $v(ab)=v(a)+v(b)$, $v(a+b)\ge\min(v(a),v(b))$, $v(0)=\infty$. For $f=\sum_ia_ix^i$, you plot all points $(i,v(a_i))$ in the plane, erect a vertical halfline above each of these, and take the convex hull of the union of these lines.
Now, an irreducible $\Bbb Q_p$-polynomial has a polygon consisting of only one nonvertical segment. Same for irreducible $\Bbb Z_p$-polynomials. But the irreducible polynomial $1-px$, whose polygon has only the two vertices $(0,0)$ and $(1,1)$, is not irreducible in $\Bbb Z_p\{x\}$, because it’s a unit, with reciprocal $1+px+p^2x^2+\cdots\,$.
You should be able to prove, from what you have been given, that the only irreducibles in $\Bbb Z_p\{x\}$ are the irreducible $\Bbb Z_p$-polynomials whose polygon has a single nonvertical segment, of nonpositive slope. The rest I leave to you.
(Maybe I should add that the Polygon is not an essential port of the proof here; in this case, it just concentrates the mind wonderfully.)