A misunderstanding about the number of points on an algebraic variety

Question

Consider the multivariate polynomial $Q(x,y) = y^2 - f(x)$ over a finite field $\mathbb{F}_q$ where $f(x) \in \mathbb{F}_q[x]$ has $m$ distinct roots in $\mathbb{F}_q$. In his book "Equations over Finite Fields", Schmidt claims that if $Q$ is absolutely irreducible then $N$, the number of points on the algebraic variety specified by $Q$, satisfies
$$|N - q| \leq (m-1) \sqrt{q}.$$ Now, what if $f(x)$ itself is irreducible? In that case we have $m=0$ and the above inequality would not make sense.

Answer 1

In this context, $m$ is the degree of $f$, and the assumption is that $f$ has $m$ distinct roots in an algebraic closure of $\mathbb{F}_q$.