Ummm. Just so you know, the same type of conclusion holds for
$$ x^2 - p y^2  $$ for prime $p,$
$$ 5 \leq p \leq 197, \; \; \; \; p \equiv 1 \pmod 4. $$
For that matter, one may switch to the forms
$$ x^2 + xy - \left( \frac{p-1}{4} \right) y^2  $$


For these forms, a number $n$ is represented if and only if $-n$ is represented. There is a solution to $x^2 - p y^2 = -1,$ a result in Mordell's book. Since every odd prime $q$ that satisfies $(p|q) = 1$ is represented by some form of the discriminant, and there is only one class of this discriminant, then all odd primes $q $ with $(q|p) = 1$ are integrally represented. Representation of the prime $2$ is a different matter, as we need $(2|p) = 1,$ so this works only when we further demand $p \equiv 1 \pmod 8.$