Assume $A$ is an $m\times n$ integral matrix (all entries are integers) with rank $m$ and $m\le n$.
I am wondering if the number of solutions to $Ax=0$ in $\{0,1,2,\dots,N\}^n$ can be bounded by $(N+1)^{O(n-m)}$, where we may assume $n(N)-m(N)$ is a constant and the integer $N$ tends to infinity. (We may also assume all $m\times m$ submatrix of $A$ has rank $m$ and even $m<n$ if it will make the bound holds.)
My feeling is that when we are in a field $F$, then the number of solutions is at most $|F|^{n-m}$. I am wondering if there is any counterpart in integral solution case.
Sure the bound holds, and in fact you can prove it the same way as you might prove it for a finite field, i.e., by decomposing $A$ and $x$ into blocks.
Without loss of generality assume the first $m$ columns of $A$ are independent. Let $A = [ B_{m \times m} \mid C_{m \times (n-m)} ]$ and $x^T = [y^T \mid z^T]$ of matching sizes s.t. $Ax = By + Cz$. Note that $B$ is invertible.
For convenience define $K = \{0, 1, \dots, N\}$. Then for any $z \in K^{n-m}$, we have
$$Ax = 0 \iff By + Cz = 0 \iff y = -B^{-1}Cz$$
So, for any $z \in K^{n-m}$, a solution for $x$ exists. However, this solution uses $y \in \mathbb{R}^m$ (or you can think of it as $\mathbb{Q}^m$, i.e. $m$-vector of rationals), so this solution might not exist for $y \in K^m$. Therefore the number of solutions for $x \in K^n$ is $\le |K|^{n-m}$.
(In fact, since $K^m$ is so much "smaller" than $\mathbb{Q}^m$, I would not be surprised if a much tighter upper bound can apply - at least, to almost all cases.)