= Solution
Use the usual positive-definite convention for <binary quadratic forms> with negative <discriminant>. Reduction by integral changes of variables of <determinant> one produces a form $ax^2+bxy+cy^2$ with $|b|\le a\le c$, and with $b\ge0$ on the boundary. For completeness, replacing $x$ by $x+ny$ makes $|b|\le a$; if then $c<a$, interchange the variables, reversing one sign to keep <determinant> one. The new positive <leading coefficient> is smaller. Repeating terminates and gives a reduced form.
For <discriminant> $-7$, $4ac-b^2=7$ and the reduction inequalities give $3a^2\le7$. Thus $a=1$, $b$ is odd and $|b|\le1$, and $c=2$. The boundary convention chooses $b=1$. Hence there is exactly one positive-definite <equivalence class>, represented by $x^2+xy+2y^2$.
For an <odd prime> $p\ne7$ represented by this form, the identity $4p=(2x+y)^2+7y^2$ shows that $-7$ is a square modulo $p$. Here $p\nmid y$, because otherwise $p\mid x$ and $p^2$ would divide the represented number. Conversely, if $-7$ is a square modulo $p$, choose odd $b$ with $b^2\equiv-7\pmod p$. Then $(p,b,(b^2+7)/(4p))$ is an integral positive-definite form of <discriminant> $-7$. By the preceding reduction it is equivalent to the given form, so its value $p$ at $(1,0)$ is represented by that form. <Quadratic reciprocity> gives $(-7/p)=(p/7)$, whose value is one precisely for $p\equiv1,2,4\pmod7$. The exceptional <primes> are represented: $2=Q(0,1)$ and $7=Q(1,-2)$. Therefore
$$
\boxed{p\text{ is represented}\iff p=7\text{ or }p\equiv1,2,4\pmod7.}
$$
Without the positive-definite convention, the first printed assertion is false: $-x^2-xy-2y^2$ also has <discriminant> $-7$ but cannot be equivalent to a positive-definite form, because an invertible <change of variables> preserves its sign.
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