An equation over a finite field Suppose $x,y,z,w \in \mathbb{F}_{q^2}$, where $q=p^k$ for some prime $p$. Consider the system of equations 
$$
\left\{
\begin{array}{l}
xy + zw = 0; \\
xy^q + yx^q + zw^q + wz^q = 0.
\end{array} 
\right.
$$
Is it possible to say something about the number of solutions (excluding the
trivial one) over $\mathbb{F}_{q^2}$? I am not that familiar with the finite field arithmetics, so sorry if this is something known.
If we substitute $xy$ from the first equation to the second, we get
$$
(xy)(x^{q-1}+y^{q-1}-z^{q-1}-w^{q-1}) = 0,
$$
but this does not make it easier for me. Thanks in advance for your help and suggestions!
UPD: By performing some computations in Mathematica, I have found that for $q=2,3,5,7$ the number of solutions takes respectively values $64,513,6625,35329$ which suggests that the number of solutions in $\mathbb{F}_{q^2}$ must be $q^3(2q^2+q-2)$, however I still don't know how to prove this in general.
 A: To just get started, consider the fact that if $(a,b,c,d)$ is a solution, then so is $(\lambda a, \lambda b, \lambda c, \lambda d)$ for every $\lambda \in \mathbb{F}_{q^{2}}$. Use this to consider two cases: first, $x = 0$ (this is rather easy; I will leave this part to you). Second, if $x \neq 0$, then you can assume $x = 1$. At the end, you can multiply the number of solutions with $x = 1$ by $q^{2}-1$ to get the total number of solutions with $x \neq 0$.
Addressing the second case, taking $x = 1$ we then have $y = -zw$ in order for the first equation to hold.  Then the second equation becomes
$$-z^{q}w^{q} - zw +zw^{q}+z^{q}w=0.$$
This simplifies to 
$$(z-z^{q})(w^{q}-w) = 0,$$
implying that either $z = z^{q}$ or $w = w^{q}$. This happens when $z \in \mathbb{F}_{q}$ or $w \in \mathbb{F}_{q}$, respectively, 
so there are $2q^{3}-q^{2}$ such solutions. 
Remember this only counts solutions with $x = 1$, so we have to multiply these by $(q^{2}-1)$ to get a total of $2q^{3}(q^{2}-1)$ solutions with $x \neq 0$.  Then there are $2q^{4}-q^{2}$ further solution with $x=0$. This gives a total of
$$(2q^{3}-q^{2})(q^{2}-1)+2q^{4}-q^{2} = q^{3}(2q^{2}+q-2)$$
solutions.
For a more technical description of this problem, you can think of your vectors $(x,y,z,w)$ as being points in a projective $3$-space $PG(3,q)$; this means that two vectors represent the same point iff they are a scalar multiple of each other (0 vector does not represent a point).  There are $q^3+q^2+q+1$ points.
Under this model you are looking for the intersection between a hyperbolic quadric (contains $(q^{2}+1)^{2}$ points) and a Hermitian surface (contains $(q^{2}+1)(q^{3}+1)$ points). Each point in this intersection then corresponds to $q^{2}-1$ nonzero vectors that satisfy both equations.
