Hopefully this will still be useful to you, or to someone in the future with a similar problem.
You have a number field $F$ and you want to find its ideal class group $C_F$. By Minkowski Bound Theorem every ideal class is represented by an ideal $I$ of norm $N(I) \leq c$ where $c$ is the Minkowski constant. So in order to find the elements of the class group, we need to find ideals of small norm in $O_F$.
There is a very important fact about ideals in rings of integers: $N(I) \in I$, so $I \mid (N(I))$. Now $N(I)$ is a natural number and can be factorised in the product of rational primes. So if we can factorise into primes all ideals $(p)$ with $p \leq c$, we will be able to find all ideals of small norm as their factors.
(Indeed prime and maximal ideals in Dedekind domains coincide.)
This is perhaps best illustrated by an example. Let $F=\mathbb{Q}(\sqrt{26})$. Then $O_F= \mathbb{Z}[\sqrt{26}]$, $n=2$, $r_2=0$ and $d_F=4\cdot 26 = 104$. The Minkowski bound is $c=\sqrt{26}<6$, so we need to find all prime ideals of norms $\leq 5$.
By Dedekind's Theorem for primes $2,3$ and $5$, we see that they factorise as
$(2) = (2, \sqrt{26})^2 =: P_2^2$ is a product of two prime ideals of norm $2$.
$(3)$ remains prime, so it has norm $9$, which is too large for our interest (ie the Minkowski Bound tells us that the same class is also represented by some ideal of smaller norm).
$(5)= (5, 1+\sqrt{26})(5, -1+\sqrt{26}) =: P_5 \cdot P_5'$ is a product of two distinct prime ideals of norm $5$.
Therefore, all ideals of norm $\leq 5$ in $O_F$ are $P_2, P_5$ and $P_5'$ and the ideal class group is generated by their classes $[P_2]$, $[P_5]$ and $[P_5']$. We have some relations between these already: $[P_5]$ and $[P_5']$ are inverses (because their product is a principal ideal) and $[P_2]$ has order $2$. Then we also observe that $(6-\sqrt{26})=P_2 \cdot P_5$ and so $[P_2] \cdot [P_5] = 1$ too and so $[P_2]=[P_5]=[P_5']$. After checking that the ideal $P_2$ is not principal, this means that the ideal class group has order $2$.
I hope this example was helpful and I'm happy to answer any further questions you may have.
Best Answer
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 $.