Let $p$ be a prime. Let $f \in \mathbb{F}\_p[x\_1, \ldots, x\_n]$ be a **homogenous** polynomial of degree $p$. Can $f$ have more than $(1-p^{-1}+p^{-2}) p^n$ zeroes in $\mathbb{F}\_p^n$?


Basic observations: The lower bound (for $n \geq 2$) is achieved by $\prod\_{c\in \mathbb{F}\_p} (x\_1 - c x\_2)$. If nonhomogenous polynomials are allowed, we can get all $p^n$ points to be zeroes, using $\prod (x\_1-c)$. It's easy to show a product of hyperplanes can't beat $(1-p^{-1}+p^{-2}) p^n$.



Most points on a degree $p$ hypersurface?
As suggested by the OP, I am making my comment an answer, as it resolves his query in the affirmative.


Theorem 2.1 (Serre, 1991), cited in *[Maximum Number of Common Zeros of Homogeneous Polynomials over Finite Fields](http://arxiv.org/pdf/1705.10185.pdf)* by Peter Beelen, Mrinmoy Datta, Sudhir R. Ghorpadel, gives an explicit formula for the maximal number of solutions in projective $(n-1)$-space of a homogeneous polynomial of arbitrary degree $d$ over finite fields of order $q$.


In particular, this maximum (which perhaps should be called the Tsfasman-Serre-Sørensen number) is: $$ dq^{n−2} + q^{n-3}+q^{n-4}+\cdots+q+1,$$
whenever $d \leq q.$


In the question, $d=p=q$ and so we have a maximum of $p^{n−1} + p^{n-3}+p^{n-4}+\cdots+p+1$ solutions in projective space. But the question was about solutions in affine space. So we scale by $\mathbb{F}\_p^\*$ then add 1 for the trivial solution to obtain:


$(p^{n−1} + p^{n-3}+p^{n-4}+\cdots+p+1)(p-1)+1=$ $$p^n-p^{n-1}+p^{n-2}=(1−p^{−1}+p^{−2})p^n,$$ as required.


