Special cases of the Stark-Heegner theorem with simple proofs The Stark-Heegner theorem states that the ring of integers of the quadratic number field $\mathbb Q(\sqrt{m})$, where $m$ is a squarefree negative integer, is a principal ideal domain, iff
$$m\in\{-1,-2,-3,-7,-11,-19,-43,-67,-163\}$$
I am looking for special cases of this theorem (including references where I can find them), which can be proven with elementary number theory or not-too-complicated algebraic number theory.
This is what I have so far:


*

*$\mathbb Q(\sqrt{m})$ is a PID for $m\in\{-1,-2,-3,-7,-11\}$ (It is easy to prove, that this rings are norm-Euclidean and that Euclidean rings are PIDs)

*$\mathbb Q(\sqrt{m})$ is a PID for $m\in\{-19,-43,-67,-163\}$ (Either prove, hat the norm is a Dedekind–Hasse norm for these rings, or that, if Euler's prime generating polynomial $x^2+x+q$ generates primes for $x=0,\dots q-1$, then  $\mathbb Q(\sqrt{1-4q})$ is a PID)

*If $m\not\equiv 1 \bmod 4$ and $m<-2$, then $\mathbb Q(\sqrt{m})$ is not a PID (Again, there are several proofs, one of them is this one)


Does anyone know any other simple result in the above spirit, particulary for proving that some rings are not PIDs? Maybe something like a proof, that there are only finitely many PIDs, maybe even an upper bound for $m$, or the fact that $m\equiv 1\bmod{8}$ does not work, or maybe some other restriction on $m$? 
Thanks!
EDIT: All Heegner numbers are prime. Maybe, there is some argument, why it won't work for composite numbers?
 A: 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

