Let $F$ be an infinite field, and $f(x, y) \in F[x, y]$ a nonzero polynomial.
Regard $f$ as a polynomial $g(y) = f(x, y) \in (F(x))[y]$. This is a polynomial in $y$, with coefficients in the infinite field $F(x)$. Since it has a finite number of distinct roots, there is a $b \in F$ such that $0 \ne g(b) = f(x, b) \in F[x]$. Now apply the result for the univariate case.
It's not the roots, it's the "$2$"!
A polynomial is irreducible over a ring if it cannot be written as a product of two non-invertible polynomials. In $\mathbb{Z}$, "$2$" is noninvertible, so $(x^2+2)2$ is an appropriately "nontrivial" factorization.
Meanwhile, over in $\mathbb{Q}$, the polynomial "$2$" is invertible, since ${1\over 2}$ is rational (proof: exercise :P). So the factoriztion $(x^2+2)2$ is "trivial" in the context of $\mathbb{Q}$, since we can always extract a factor of $2$ from any polynomial.
EDIT: Think of it this way: saying that a polynomial is irreducible over a ring means it has no "nontrivial" factorizations. Now, when we make the ring bigger (e.g. pass from $\mathbb{Z}$ to $\mathbb{Q}$) two things happen:
So even though your first instinct might be "polynomials will only go from "irreducible" to "reducible" as the ring gets bigger," actually the opposite can happen!
In fact, here's a good exercise:
Can you find a polynomial $p\in\mathbb{Z}[x]$ which is irreducible over $\mathbb{Z}$ but reducible over $\mathbb{Q}$?
Note that the definition of reducibility over a field may sound different:
For $F$ a field, a polynomial $p\in F[x]$ is irreducible if $p$ cannot be written as the product of two nonconstant polynomials.
But this is actually equivalent to the definition I gave above, in case we're over a field: the noninvertible elements of $F[x]$ are precisely the nonconstant polynomials!
Best Answer
Versions of this question have been much studied, but it's not obvious what the correct keywords are. In this case you want to look at algebraic geometry, particularly the study of algebraic curves, over finite fields. The relevant theorem in this subject is the Hasse-Weil bound, which in turn massively generalizes to the Weil conjectures.
The Hasse-Weil bound implies that if $f$ is irreducible and a certain nonsingularity condition is met (actually I am not sure if this is necessary), the number of solutions is something like (but not quite)
$$p \pm 2g \sqrt{p}$$
where $g = \frac{(r-1)(r-2)}{2}$ by the genus-degree formula. This gives a bound which is worse than your bound if $r$ is large compared to $p$ but which is substantially better if $p$ is large compared to $r$, so which bound is better depends on which regime you care about.
Of course, if $r > p$ then an even better bound than $rp$ is $p^2$...
The Hasse-Weil bound should conjecturally be tight in a suitable sense (in particular, asymptotically in $p$) by a suitable generalization of the Sato-Tate conjecture. On the other hand, in general your bound can be tight, since $f$ may be a product of distinct linear factors. If $f$ is reducible then you should factor $f$ and apply the Hasse-Weil bound separately to each of its factors.