The first thing to note is that ideal classes are different from ideals, we say $I$ is in the class of $J$ if $\exists (a),(b)$ with $(a)I=(b)J$. With this in mind, consider $(4\sqrt{-55})(5,\sqrt{-55})=(20\sqrt{-55},-220)=(\sqrt{-55})$. Now consider the ideal $(2,\frac{1+\sqrt{-55}}{2})$. I claim this is in the same class as $(5,\sqrt{-55})$.
Indeed, consider $$(10\sqrt{-55})(2,\frac{1+\sqrt{-55}}{2})=(20\sqrt{-55},5\sqrt{-55}+275)=(\sqrt{-55})$$
This answers your original question.
To get an ideal of norm $<5$, you want to consider ideals with norm $2$ or $3$. The way we do this is to determine whether $(2) , (3)$ are ramified in $K=\mathbb{Q}(\sqrt{-55})$.
Since $-55\equiv 1\pmod{4}$ we know that Disc($K)=-55$, and since neither $2$ nor $3$ divides 55, we see that neither prime is ramified. Thus, it remains to decide whether they are split completely or inert.
$(3)$ remains a prime in $K$, since $3$ is not a quadratic residue modulo $-55$, so we can discount ideals of norm $3$.
$(2)\equiv1\pmod{8}$, so we know $(2)$ splits completely in $K$. The ideal I=$(2,\frac{1+\sqrt{-55}}{2})$ is a prime ideal of norm $2$, since $(2,\frac{1+\sqrt{-55}}{2})(2,\frac{1-\sqrt{-55}}{2})=(2)$. A computation shows that $I^2$ is not principal, and indeed, that $I^2\neq \bar I$. Hence, $I$ is not of order $3$. We compute either $I^3$, to give us $\bar I $, or $(I^2)^2$, to give us a principal ideal. Either way, we see that $I$ has order 4, and the class group is henceforth $\mathbb{Z}/4\mathbb{Z}$.
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
In general, use the definition of equivalence and show that there exist elements $\alpha$ and $\beta$ with $\alpha {\mathfrak a} = \beta{\mathfrak b}$.
In your example we have $(2,\sqrt{10})(3,1-\sqrt{10}) = (2 + \sqrt{10})$ and $(2,\sqrt{10})(3,1+\sqrt{10}) = (2 - \sqrt{10})$, from which it follows that the two ideals in question belong to the same ideal class. If you divide the first equation by the second and clear denominators, then you get $$ (2 - \sqrt{10})(3,1-\sqrt{10}) = (2 + \sqrt{10})(3,1+\sqrt{10}) , $$ which gives you the elements $\alpha$ and $\beta$ mentioned above. The ideals in the displayed equation both equal $3(2,\sqrt{10})$, by the way (look at their prime ideal factorization).
How do we find these equations? Well, we have ideals of norms $2$ and $3$, so to find whether they are in the same class we look for elements whose norms are $\pm 6$, $\pm 12$, $\pm 18$ etc.