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 $.
The latter factorizes mod 2 as $x(x+1)$, so $(2)=\left(2,\frac{1+\sqrt{−199}}{2}\right) \left(2,\frac{3+\sqrt{−199}}{2}\right)$, neither of the factors being principal (and hence both of order 2)
Why do you think that these prime ideals being nonprincipal implies that they have order 2?
Labeling the prime ideals, say $(2) = \mathfrak{p}\mathfrak{p}'$. If the order of $\mathfrak{p}$ is 2, then necessarily $\mathfrak{p} = \mathfrak{p}'$ (hint: you can use that they have the same norm and are equal in the class group). Can you show that isn't the case?
In fact, $\mathfrak{p}^k$ isn't principal for $k = 1,2,\dots,8$, but $\mathfrak{p}^9 = \left(\frac{43\pm\sqrt{-199}}{2}\right)$ (sign depending on choice of $\mathfrak{p}$). Hence the order of $\mathfrak{p}$ in the class group is 9.
Best Answer
For any non-zero ideal $I$ there is an element $a\in I-0$ such that $N(a)\le 10 N(I)$. So the inverse of $I$ in the class group is represented by an ideal of norm $\le 10$. Enumerating all those ideals of norm $\le 10$ you'll find that they are either principal or in the class of $(2,1+\sqrt{-5})$ (enumerating the prime ideals of norm $\le 10$ is enough).
For the construction of $a$: let $m=\lfloor \sqrt{N(I)}\rfloor+1 $ and consider the set $S=\{u+v\sqrt{-5}, u,v \in 1\ldots m\}$. We have listed $m^2>N(I)$ elements of $\Bbb{Z}[\sqrt{-5}]$ so there must be two elements of $S$ that are the same in $\Bbb{Z}[\sqrt{-5}]/I$ ie. $(u_1-u_2)+(v_1-v_2)\sqrt{-5}\in I-0$ with $|u_1-u_2|\le m,|v_1-v_2|\le m$.