There are tons of easy cases. If $m$ is even, or if $m \equiv -1 \bmod 4$, then $2$ ramifies in $\mathbb{Q}(\sqrt{m})$. But $2$ is not of the form $a^2+|m| b^2$, except for $m=-1$ and $m=-2$. So we can limit ourselves to $m \equiv 1 \bmod 4$. As pointed out above, we can also assume $m \not \equiv 1 \bmod 8$. So we are down to $m \equiv -5 \bmod 8$. This means that the ring of integers in $\mathbb{Q}(\sqrt{m})$ is $a+b(1+\sqrt{m})/2$, with norm function $a^2 + ab + \frac{|m|+1}{4} b^2$. (Of course, $m$ is negative, but I think a lot of these formulas are more readable with $|m|$ instead of $-m$ because it makes the sign immediately visually clear.)
In particular, if $b \neq 0$, the norm $N(a+b(1+\sqrt{m})/2) \geq \frac{|m|+1}{4}$.
First of all, this explains why $m$ must be prime. If $m$ is an odd, square-free composite, then there is a prime $p$ dividing $m$ with $p \leq |m|/5$. This prime $p$ ramifies, so there is a prime ideal $\pi$ with $N(\pi) = p$. Since $p \leq m/5 < \frac{|m|+1}{4}$, the ideal $\pi$ can't be principal.
Also, suppose that $p$ is an odd prime with $\left( \frac{m}{p} \right) =1$ and $p < (|m|+1)/4$. Then, $p$ splits into $\pi \bar{\pi}$ and the same argument as the above paragraph shows that $\pi$ can't be principal.
Summary Any counterexample must have $|m|$ a prime, $m \equiv 5 \bmod 8$, and must have $\left( \frac{m}{p} \right) = -1$ for $p < \frac{|m|+1}{4}$.
I can't find any $m$ more negative than $-163$ which passes this test, searching through the first $50,000$ primes. Nothing even comes close. Define $r(m)$ to be the least odd prime $p$ for which $\left( \frac{m}{p} \right) = 1$. A PID would have $r(m)/|m| > 0.25$; the largest value I can find after $163$ is $r(-193)/193 = 11/193 \approx 0.057$. In case you want to do some computations yourself, here is a bit of Mathematica code:
FirstSplit[m_] :=
(i = 2; While[KroneckerSymbol[m, Prime[i]] == -1, i++]; Prime[i])
SplitRatio[m_] := FirstSplit[-m]/m
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 $d \in \{1,2,3,7,11,19,43,67,163\}$
Best Answer
I cannot give a comprehensive list, nor point you at published articles/books. My "go to" -source for bits like these is the LMF-database.
Caveat: I am not very experienced in interpreting the database output.