$F$ is a finite field and $p \in F[x_1, x_2, \ldots ,x_n]$ is a polynomial
If $p$ restricts to $0$ on a hyperplane $q = a_0 + a_1 x_1 + \cdots + a_n x_n = 0$ in $F^n$ then does it follow that $q$ divides $p$?
By 'restricts to $0$' I mean if we parameterise $q$ by $n-1$ paramteters $t_i$ and compute the restriction of $p$ to $q$ (i.e. a polynomial in $F[t_1, \ldots, t_{n-1}]$ where $t_i$ are our parameters) this restriction is the zero polynomial.
Suppose $n > 1$.
Consider the polynomial $p(x_1, \dots, x_n) = x_1^2 - x_{1}$ over the field $F = \mathbf{F}_{2}$. (So I might have written $-1 = 1$.)
Clearly $p$ evaluates to zero on any element of $F^{n}$. But $p$ is not divisible by, say, $q = x_2$.
The above was a solution to the problem as I understood it by reading a previous version, assuming that $p$ evaluates to zero on the hyperplane.
In the current formulation, I believe the answer to be positive.
Suppose WLOG $a_{n} \ne 0$, and do a change of variables so that the hyperplane is $q = x_{n} = 0$. (And the hyperplane is parametrized by $x_1, \dots, x_{n-1}$ in the obvious way.)
Now consider $p$ as a polynomial $p' \in K[x_{n}]$ in the variable $x_n$, with coefficients in $K = F[x_1, \dots, x_{n-1}]$.
If I understand correctly the assumption, this says that $p'(0) = 0$, that is, $p'$ has no constant term when regarded as a polynomial in $K[x_{n}]$, that is, it is a multiple of $q = x_{n}$, as required.