A polynomial $p(x,y)$ that is never $0$ if $y\neq 0$ I would like to find the best proof of the following fact:
If a polynomial $p\in k[x,y]$ (where $k$ is an algebraically closed field) is such that $p(a,b)\neq 0$ on the set $\{(a,b):b\neq 0\}$, then in fact $p\in k[y]$ (that is, $p$ does not depend on $x$). I do know how to prove this using an argument with a Vandermonde matrix, but I feel that there should be a more direct way to do it.

As an amusing complement to anomaly's excellent answer, notice that the polynomial $p(x,y)=a_0(y)$ being nonzero for $y\neq0$ is necessarily of the form $f(x,y)=cy^n$ where $c\in k^*$ and $n\in \mathbb N$.
Notice also that the theorem is false for non algebraically closed fields, as witnessed by the polynomial $x^2+1\in \mathbb R[x,y]$ .
