You may be able to push this kind of argument through:
Suppose 5 divides both $y+10\sqrt{-2}$ and $y-10\sqrt{-2}$. Then $5\mid x^3$, so $5^3\mid x^3$, so $5^2$ must divide one or the other of $y\pm10\sqrt{-2}$, but neither of the numbers $(y\pm10\sqrt{-2})/25$ is in ${\bf Z}[\sqrt{-2}]$. Hence, 5 is not a common divisor.
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
$I^k$ divides $I^g$, so $(1+\sqrt{-d})=I^g \subset I^k = <a>$. Hence, $a=\pm 1$, which implies that $I^k=\mathbb{Z}[\sqrt{-d}]$. Then $I^g=\mathbb{Z}[\sqrt{-d}]$ as well. However, $I^k=(1+\sqrt{-d})$ and we get a contradiction as required.