I know that it is a well-known theorem that for a fixed $n$ there are only finitely many $D<0$ such that $h(\mathbb{Q}(\sqrt{D}))=n$. However, I was told that this is not true for the case when $D>0$. I think I have an idea (although it is most likely wrong), but I am curious as to why it might be wrong. In D.A. Buell's "Binary Quadratic Forms" Theorem $6.19$ states the following
a) If the discriminant $\Delta$ of the quadratic field $\mathbb{Q}(\sqrt{\Delta})$ is negative, then the class group and the narrow class group are isomorphic
b) If the discriminant $\Delta$ of the quadratic field $\mathbb{Q}(\sqrt{\Delta})$ is positive and a solution exists to the equation $$x^2-\Delta y^2=-4$$ then the class group and the narrow class group are isomorphic
c) If the discriminant $\Delta$ of the quadratic field $\mathbb{Q}(\sqrt{\Delta})$ is positive, and no solution exists to the equation $$x^2-\Delta y^2=-4$$ then the class group consists of the subgroup of squares of the narrow class group.
Furthermore, I saw this post for showing that the size of the narrow class group is a multiple of class number. Thus, shouldn't it be the case that if I fix some number $n$, and I wish to show that there are only finitely many positive $D>0$ such that $h(\mathbb{Q}(\sqrt{D}))=n$, it would be sufficient to show that there are only finitely many classes of binary quadratic forms of a given discriminant which is true? Perhaps, there is something wrong with the case that $$x^2-\Delta y^2=-4$$ has no solutions as then the class group is isomorphic to the squares of the narrow class group. Any help on where my confusion might lie would be appreciated.