Let $\alpha=\frac{1+\sqrt{-35}}{2}$, then in fact, $(3)=Q\overline{Q}=(3,\alpha)(3,\alpha+2)$. In the class group then, $[Q]+[\overline{Q}]=[\mathcal{O}_K]$, so $[Q]$ and $[\overline{Q}]$ are inverses. So the class group is generated by $Q$, and we just need to show that $Q^2$ is principal.
The minimal polynomial of $\alpha$ is $f(x)=x^2-x+9$, and recall that $f(\beta)=\text{Nm}(\beta-\alpha)$. In particular, $f(0)=9=3^2$. The element $(\alpha)$ is sent to $0$ under the evaluation $\alpha\mapsto 0$, corresponding to the ideal $Q$, so $(\alpha)=Q^2$, and the order of $[Q]$ divides $2$. You've already shown that $Q$ can't be principal, so we're done.
From your calculation, you know the following facts about the class number $ h = h_{\mathbb{Q}(\sqrt{-17})} $ already
$ h > 1 $, because $ \mathfrak{p}_2 = (2, 1 + \sqrt{-17}) $ is a non-principal ideal
$ h \leq 5 $, because you have found 5 ideals with norm less than the Minkowski bound.
We have $ h = 2, 3, 4, \text{ or } 5 $. But to pin down $ h $, you need to do more work...
Step 1: From your computation of $ \mathfrak{p}_2 = (2, 1 + \sqrt{-17}) $ being an ideal of norm 2, you probably have found that $ (2) = \mathfrak{p}_2^2 $ since $ -17 \equiv 3 \pmod{4} $. Or just directly check this by computing the product $ \mathfrak{p}_2^2 $.
Because $ (2) $ is principal, this shows that in the class group, $ [\mathfrak{p}_2] $ is an element of order 2. But what do you know about the order of an element in a finite group? It must divide the order of the group. So which possibilities for $ h $ can we now eliminate?
Step 2: We can ask a similar question about $ \mathfrak{p}_3 = (3, 1 + \sqrt{-17}) $: is $ \mathfrak{p}_3^2 $ principal? If it was, then it would be an ideal of norm $ 3 \times 3 = 9 $. So it would be generated by an element $ \alpha $ with norm $ \pm 9 $. But solving $ N(\alpha) = x^2 + 17y^2 = 9 $ shows $ \alpha = \pm 3 $. But when you have factored $ (3) $ to find $ (3) = \mathfrak{p}_3 \widetilde{\mathfrak{p}_3} $, you can also say $ \mathfrak{p}_3 \neq \widetilde{\mathfrak{p}_3} $. This means that by the unique factorisation of ideals in number rings, $ \mathfrak{p}_3^2 \neq (3) $, so it cannot be principal.
(Alternatively: compute that $ \mathfrak{p}_3^2 = (9, 1 + \sqrt{-17}) $. If this is going to equal $ (3) $, then we must have $ 1 + \sqrt{-17} \in (3) $, but clearly $ 3 \nmid 1 + \sqrt{-17} $, so this is not possible. Again conclude $ \mathfrak{p}_3^2 $ is not principal.)
This means that in the class group $ [\mathfrak{p}_3] $ has order $ > 2 $. This shows that $ h > 2 $ because the order of an element must divide the order of the group.
Conclusions: You have enough information now to conclude that $ h = 4 $, agreeing with Will Jagy's answer. Step 1 shows that $ h = 2, 4 $, and step 2 shows that $ h = 4 $, so we're done.
In fact we also know the structure of the class group $ \mathcal{C}(\mathbb{Q}(\sqrt{-17})) $. It is a group of order $ h = 4 $, so either $ \mathcal{C}(\mathbb{Q}(\sqrt{-17})) \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \text{ or } \mathbb{Z}_4 $. But since $ \mathfrak{p}_3 $ has order $ > 2 $, it must have order 4. This shows that $ \mathcal{C}(\mathbb{Q}(\sqrt{-17}) \cong \mathbb{Z}_4 $.
Best Answer
Completing the program suggested by reuns’s argument:
Let $m$ be such a discriminant and assume that $m \neq 1 \pmod{4}$. Then the ring of integers of $K$ is $\mathbb{Z}[\sqrt{m}]$, so $K$ is ramified at $2$. If $K$ has class number one, the ideal $I$ whose square is two is principal of generator $\alpha$, and the norm of $\alpha$ is then a positive integer with absolute value $2$, so $2$ is a norm. Thus $x^2-my^2=2$ has a solution. As $m <0$, this implies $m \in \{-1,-2\}$.
We’re going to do this again with other primes. So now we always assume that $m=1\pmod{4}$, ie $m=1-4A$, so that the ring of integers is $\mathbb{Z}[\omega]$ with $\omega=\frac{1+\sqrt{m}}{2}$.
Assume that $p$ is a prime dividing $m$, so $p>2$ is coprime to $A$ and $4A-1=Bp$: then $K$ is ramified at $p$, and the generator $\alpha$ of the ideal whose square $(p)$ has, as above, norm $p$. But write $\alpha=x+y\omega$, then the norm of $\alpha$ is $x^2+xy+Ay^2=p$. This can be re-written as $(x+y/2)^2+\frac{(4A-1)}{4}y^2=p$. So if $B>4$, then $\frac{(4A-1)y^2}{4} > py^2 \geq p$, hence a contradiction. As $B$ is odd, $B$ is one or three.
But by doing the same reasoning with $3|m$ (if $B=3$), we see that $p = 3$, so $m=-9 \neq 1\pmod{4}$. Therefore $m=-p$, with $p = 3\pmod{4}$ prime.
We can assume $p \geq 23$. Then, if $q<p/4$ is an odd prime, $q$ splits in $K$ iff $-p$ is a square mod $q$. But if $q$ splits in $K$, then, as above, $q$ must be a norm, ie we must have integer solutions to $(x+y/2)^2+py^2/4=q$. As $q < p$, we must have $y=0$, so $q=x^2$, which is absurd.
So $-p$ is never a square modulo any odd prime smaller than $p/4$. In particular, we must have $p=1\pmod{3}$ (so $p=7\pmod{12}$, disqualifying $23$, so we assume $p\geq 31$), $p=2,3\pmod{5}$ (disqualifying $31$, so we can assume $p> 43$), $p=1,2,4\pmod{7}$.
As $p\geq 43$ is congruent to $7$ mod $12$, if $p \leq 163$, then $p \in \{55,67,79,91,103,115,127,139,151,163\}$. Of them, we rule out $55,91,115$ for being non-prime, $67,163$ for being in the problem statement, $79,139,151$ by the criterion mod $5$, and only $127$ remains.
We rule out $127$ by the criterion mod $11$ – as $-127$ is a square mod $11$.
Anyway, the four congruence conditions above make for six congruence classes mod $420$, so at most $30$ integers less than $2000$.
We compute that these congruence classes are $43,67,127,163,247,403$, so we get the integers $247,403$ and everyone plus $420$, plus $840$, plus $1260$, plus $1680$.
Let’s add a few more criteria, with $q=11,13,17$: $p=1,3,4,5,9\pmod{11}$, $p=2,5,6,7,8,11\pmod{13}$.
By the mod $11$ criterion, we see that only remain $43+420k$ ($1 \leq k \leq 3$), $67+420k$ ($1 \leq k \leq 4$), $127+420k$ ($k=3,4$), $163+3 \cdot 420$, $247+k\cdot 420$ ($k=0,2$), $403+k\cdot 420$ ($k=1,4$).
Combining with a look mod $13$, only the integers $43+k\cdot 420$ ($k=1,2,3$), $67+4\cdot 420$, $247+2\cdot 420$ remain.
Now, we compute (by hand!) that $-463, -883, -1087,-1747$ are squares mod $17$, while $-1303$ is a square mod $23$, which rules them out.