This is an expository note filling in the background between Steven Sam's comment and Felipe Voloch's answer.
If $X$ is a smooth projective variety, then the Weil conjectures (now theorems) describe the zeroes and poles of the zeta function in terms of the cohomology of $X$, and the action of Frobenius on it. In particular, the poles on the circle $|u|=1/q$ are the reciprocals of the eigenvalues of Frobenius acting on $H^2(X, mathbb{Q}_{ell})$.
In your example, $H^2$ is two dimensional. Over the algebraic closure $overline{F_q}$, your variety is isomorphic to $mathbb{P}^1 times mathbb{P}^1$. $H^2$ is spanned by the two classes $mathbb{P}^1 times { mbox{point} }$ and $ { mbox{point} } times mathbb{P}^1$.
If $-1$ is a square, Frobenius acts on this two dimensional vector space by multiplication by $q$, so you get a double pole at $1/q$. If $-1$ is not a square, then Frobenius multiplies by $q$ and switches the two generators. So the eigenvalues are $q$ and $-q$.
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