Here's what I think is a nice way to find a few PIDs that aren't Euclidean. It's not quite elementary, but if you know a bit about number fields I think it's a lot easier and nicer than the normal drudge.
Let $K=\mathbb{Q}(\sqrt{-d})$ for $d>3$ squarefree, with ring of integers $\mathcal{O}_K$. Then the only units in $\mathcal{O}_K$ are $\pm1$.
Suppose $\mathcal{O}_K$ is Euclidean with Euclidean function $\varphi$. Then take $x \in \mathcal{O}_K\setminus\{0,\pm1\}$ with $\varphi(x)$ minimal. By definition, any element of $\mathcal{O}_K$ can be written in the form $px+r$ where $\varphi(r) < \varphi(x)$, so it must be that $r \in \{0,\pm1\}$, i.e. $|\mathcal{O}_K/(x)|$ is $2$ or $3$. In other words $\mathcal{O}_K$ has a principal ideal of norm $2$ or $3$.
So now we know that if $K = \mathbb{Q}(\sqrt{-d})$ has class number one, where $d>3$ is squarefree*, $K$ is a non-Euclidean PID if there are no elements in $\mathcal{O}_K$ of norm $\pm2$ or $\pm3$. As $K$ is a PID (and degree $2$ over $\mathbb{Q}$), this is equivalent to saying that $2$ and $3$ are inert. To find some examples then:
If $d = 3\pmod{4}$, $\mathcal{O}_K = \mathbb{Z}[\frac{1+\sqrt{-d}}{2}]$, the minimal polynomial of $\frac{1+\sqrt{-d}}{2}$ over $\mathbb{Q}$ is $f_d(X)=X^2-X+\frac{1+d}{4}$. Applying Dedekind's criterion gives that $d$ works provided that $f_d(X)$ is irreducible $\pmod{2}$ and $\pmod{3}$. This then gives that $d = 19$ works (which is the usual example), but also shows that $d = 43,67$ or $163$ work as well (I think!).
*It can be shown that this implies $d \in \{1,2,3,7,11,19,43,67,163\}$
With $d<0$, the ring of integers of $\mathbb Q(\sqrt d)$ is a PID exactly when
$$d= −1, −2, −3, −7, −11, −19, −43, −67, −163$$
Checking that these rings are PIDs is not too hard, but checking that no other values give an integer ring that is a PID requires some heavy machinery - see the class number 1 problem.
Best Answer
The ring $\Bbb Z[\sqrt{5}]$ is not a UFD, but the ring of integers $\Bbb Z[\frac{1+\sqrt{5}}{2}]$ of the number field $\Bbb Q(\sqrt{5})$ is a Dedekind ring, which is a PID and hence is factorial.
More references (in addition to your links, with answers concerning this topic):
For which $d$ is $\mathbb Z[\sqrt d]$ a principal ideal domain?
Edit: The proof that $K=\Bbb Q(\sqrt{5})$ has class number $1$ is very easy, because the Minkowski bound satisfies $B_K<2$, so that the class number is $1$. More precisely, $$ B_K=\frac{\sqrt{5}}{2}=1.11803398874989\cdots $$ The formula in general is $$ B_K=\frac{n!}{n^n}\left(\frac{4}{\pi} \right)^s\sqrt{|d_K|} $$ where $n$ is the degree of the extension (here $n=2$), $d_K$ the discriminant.