$\newcommand{\char}{\operatorname{char}}$Given a finite field $F_q$ with $q\equiv 1 \bmod 3$ and $\char(F_q)>3$, I need to figure out how many isomorphism classes of elliptic curves $E/F_q$ have a square discriminant $\Delta:=-16(4a^3+27b^2)$.
Since $\char(F_q)>3$, we can consider short Weiestrass curves only ($y^2=x^3+ax+b$) where isomorphisms are given by $$(a,b)\mapsto (\lambda^4a, \lambda^6b),~\lambda\in F_q^*$$ and we can split isomorphism classes into four "superclasses" depending on whether $a=0$ and/or $b=0$.
Superclass 1: $a=b=0$
There are no elliptic curves here since it is singular.
Superclass 2: $a=0,~b\neq0$
The action of $F_q$ on the set of elliptic curves via $\lambda \ast b := \lambda^6 b$ has orbits of size $(q-1)/6$ (since both 2 and 3 divide $p-1$ by assumption). Since there are a total of $q-1$ curves here, this means there are 6 isomorphism classes in this superclass, and all of them have a square discriminant since $\Delta = -432b^2$ with $-432$ being a square since $q\equiv 1\bmod 3$.
Superclass 3: $a\neq0,~b=0$
By a similar argument on the orbits of $\lambda\ast a := \lambda^4 a$, if $q\equiv 3 \bmod 4$ then there are 2 isomorphism classes, whereas if $q\equiv 1\bmod 4$ then there are 4 isomorphism classes. Since $\Delta = -8^2a^3$, exactly half of the classes in either case have a square discriminant (this follows since all isomorphism classes have the same size and $a^3$ is a square half of the time).
Superclass 4: $a,b\neq0$
Each isomorphism class has size $(p-1)/2$ (the orbit size of $\lambda\ast(a,b):=(\lambda^4a,\lambda^6b)$), so it would suffice to find the raw number of solutions to $-16(4a^3+27b^2)=\delta^2$. Via a change of variables $(a',b',\delta')=(-a,\frac{3\sqrt{-3}}{4}b,\frac{\delta}{8})$, where $\sqrt{-3}$ exists since $q\equiv 1\bmod 3$, this can be simplified to $$a^3+b^2=\delta^2,$$ so counting the number of points in this 2D surface (subject to $a,b\neq 0$) would conclude the task. I know there are some bounds for this, but I have been trying to get an exact solution to it without success.
