Given a finite field $F_q$ with $q\equiv 1 \mod 3$ 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 $$(a,b)\mapsto (\lambda^4a, \lambda^6b)$$ 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 $q\equiv 1\mod 3$.

Superclass 3: $a\neq0,~b=0$

By a similar argument on the orbits of $\lambda\ast a := \lambda^4 a$, if $q\equiv 3 \mod 4$ then there are 2 isomorphism classes, whereas if $q\equiv 1\mod 4$ then there are 4 isomorphism classes. Since $\Delta = -8^2a^3$, exactly half of the classes in either case have a square discriminant.

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 $(a',b',\delta')=(-a,\frac{3\sqrt{-3}}{4}b,\frac{\Delta}{64})$, where $\sqrt{-3}$ exists since $q\equiv 1\mod 3$, this can be simplified to $$a^3+b^2=\delta^2,$$ so counting the number of points in this 2D surface 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.