= Solution
Write an integral <binary quadratic form> as
$$
[a,b,c](x,y)=ax^2+bxy+cy^2,
\qquad
d=b^2-4ac<0.
$$
A positive definite form is a <reduced positive definite binary quadratic form> when
$$
|b|\leq a\leq c,
$$
with $b\geq0$ on the boundary $|b|=a$ or $a=c$. The <class number of a negative discriminant> $h(d)$ is the number of <proper equivalence of binary quadratic forms>[proper equivalence classes] of primitive positive definite integral forms of discriminant $d$.
Every such class has a reduced representative. For a reduced form,
$$
4a^2\leq4ac=b^2+|d|\leq a^2+|d|,
$$
so
$$
a\leq\sqrt{\frac{|d|}{3}}.
$$
There are only finitely many possible integers $a$ and $b$, and then
$$
c=\frac{b^2-d}{4a}
$$
is determined. Hence $h(d)<\infty$. There is at least one class: the principal form is
$$
[1,0,-d/4]\quad\hbox{if }d\equiv0\pmod4,
$$
and
$$
[1,1,(1-d)/4]\quad\hbox{if }d\equiv1\pmod4.
$$
Thus $1\leq h(d)<\infty$, in agreement with the <enumeration of reduced binary quadratic forms>.
Now let $q\equiv3\pmod8$ be prime. The two forms
$$
f_0=[1,0,2q],
\qquad
f_1=[2,0,q]
$$
are primitive, positive definite, reduced, and have discriminant $-8q$. They are not <proper equivalence of binary quadratic forms>[properly equivalent]: $f_0$ represents one, while the least positive value of $f_1$ is two. Therefore
$$
\boxed{h(-8q)\geq2.}
$$
Assume henceforth that $h(-8q)=2$, so these are the only two classes. If
$$
p=x^2+2qy^2
$$
for a prime $p>q$, then $x$ is odd and
$$
p\equiv x^2+6y^2\equiv
\begin{cases}
1\pmod8,&y\text{ even},\\
7\pmod8,&y\text{ odd}.
\end{cases}
$$
Moreover $p\equiv x^2\pmod q$, so $p$ is a <quadratic residue> modulo $q$. This proves the necessity.
Conversely, suppose $p\equiv\pm1\pmod8$ and $p$ is a quadratic residue modulo $q$. By <quadratic reciprocity>, these two conditions imply
$$
\left(\frac{-8q}{p}\right)=1.
$$
Indeed, for $p\equiv1\pmod8$ both relevant signs are positive, while for $p\equiv7\pmod8$ the signs from $(-8/p)$ and reciprocity with $q\equiv3\pmod4$ cancel. Choose an even integer $b$ such that
$$
b^2\equiv-8q\pmod{4p}.
$$
Then
$$
[p,b,c],
\qquad
c=\frac{b^2+8q}{4p},
$$
is a primitive positive definite integral form of discriminant $-8q$ and represents $p$. It belongs to one of the two classes $f_0,f_1$.
The second form cannot represent such a prime. If
$$
p=2x^2+qy^2,
$$
then $y$ is odd and, since $q\equiv3\pmod8$,
$$
p\equiv
\begin{cases}
3\pmod8,&x\text{ even},\\
5\pmod8,&x\text{ odd}.
\end{cases}
$$
This contradicts $p\equiv\pm1\pmod8$. Hence $[p,b,c]$ lies in the principal class $f_0$, and <proper equivalence of binary quadratic forms> preserves represented integers. Therefore
$$
\boxed{
p=x^2+2qy^2
\quad\Longleftrightarrow\quad
p\equiv\pm1\pmod8
\ \hbox{and}\
\left(\frac pq\right)=1.}
$$
This is the <prime representation when h of minus eight q equals two>.
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