Let me elaborate a bit on some of the ideas implicit in Jyrki Lahtonen's very slick answer. In particular, I will give a concrete sense in which the vanishing set of a polynomial is "small", analogous to how the vanishing set of a polynomial over $\mathbb{R}$ has measure zero (in fact, this argument can be used to prove that measure-theoretic fact).
Specifically, we define by induction on $r$ what it means for a subset $V\subseteq F^r$ to be "small". For $r=0$, we say $V\subseteq F^0$ is small if it is empty. By induction, we then say that $V\subseteq F^r$ is small iff for all but finitely many $a\in F$, $\{b\in F^{r-1}:(b,a)\in V\}$ is a small subset of $F^{r-1}$. So for instance, a subset of $F^1$ is small iff it is finite, and a subset of $F^2$ is small iff all but finitely many of its vertical fibers are finite.
Now I claim that if $f\in F[x_1,\dots,x_r]$ is any nonzero polynomial, then the set $V=\{v\in F^r:f(v)=0\}$ is a small subset of $F^r$. We prove this by induction on $r$; the case $r=0$ is trivial. Supposing $r>0$, think of $f$ as a polynomial in $x_1,\dots,x_{r-1}$ with coefficients in $F[x_r]$. Since $f\neq 0$, at least one of these coefficients in $F[x_r]$ is nonzero. There are thus only finitely many values which can be substituted for $x_r$ which make all the coefficients $0$. Thus for all but finitely many $a\in F$, the polynomial $f_a(x_1,\dots,x_{r-1})=f(x_1,\dots,x_{r-1},a)$ is nonzero. By induction, the set $\{b\in F^{r-1}:(b,a)\in V\}=\{b\in F^{r-1}:f_a(b)=0\}$ is small for all but finitely many $a$. Thus $V$ is small.
Now finally note that the union of two small sets is small (easy by induction on $r$) and that if $F$ is infinite, then $F^r$ is not small as a subset of itself (again, easy by induction on $r$). Thus if $f$ and $g$ are nonzero polynomials, the union of their vanishing sets is small and in particular is not all of $F^r$. (Alternatively, as in the other answer, the union of their vanishing sets is the vanishing set of $fg$ and hence is small.)
