There are two parts to showing a Euclidean domain or a PID are UFDs: (i) existence of an irreducible factorization for every nonzero nonunit and (ii) essential uniqueness of the irreducible factorization (any two use the same number of irreducible factors and the irreducibles that occur in both factorizations can be matched termwise up to multiplication by a unit).
To prove (ii), the key point is that every irreducible element is a prime element, and to prove that you need to be able to write $px + ay = 1$ for any irreducible $p$ and element $a$ where $p \nmid a$; the nondivisibility implies (since $p$ is irreducible) that the only common factors of $p$ and $a$ are units, so Euclid's algorithm in a Euclidean domain lets you algorithmically solve $px + ay = 1$ for some $x$ and $y$. In a PID you'd instead observe that the ideal $(p,a)$ has to be $(1)$.
To prove (i) is a major distinction between Euclidean domains and PIDs. I compare approaches for each (you can indeed handle the Euclidean case without using the proof for PIDs in disguise) as Theorems 4.2 and 4.3 at http://www.math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf. You need to read Sections 2 and 3 first to see why I mean about being able to assume the "$d$-inequality" holds. A Euclidean domain does not have to have its "norm" function $d$ be totally multiplicative or satisfy $d(a) \leq d(ab)$, but you can always adjust the "norm" function to fit that inequality all the time. Of course in $\mathbf Z$ and $F[x]$ that inequality is true, so you save some time when proving those rings are UFDs compared to a general Euclidean domain.
The bottom line is that you definitely do not need to introduce the machinery of PIDs in order to prove rings like $\mathbf Z$ or $F[x]$ have unique factorization. After all, unique factorization in those types of rings as well as in $\mathbf Z[i]$ was known (say, to Gauss) long before there was a concept of PID.