Are there infinitely many $D$ such that $Cl_K[2] \cong \mathbb{Z} / 2\mathbb{Z}$?
It's well-known (and straighforward to show) that $Cl_K[2]$ has order $2^{r-1}$ where $r$ is the number of prime factors of the discriminant. Since it's clear that there are infinitely many fundamental discriminants with precisely two prime factors, the answer is "yes".
(EDIT: I added a quotation to clarify what exactly I am answering, since the original post hints at several distinct questions with rather different answers.)