To start with, I would like to give an example. Suppose $f(n)=\frac{n(n-1)}{2}$ and $m=2^k$ a two power. Then the map $$ \mathbb{Z}\rightarrow\mathbb{Z}/m\mathbb{Z}, \quad n\mapsto f(n)\mod m, $$ is surjective. It is because if we choose $c\in\mathbb{Z}/m\mathbb{Z}$, then $$ f(n)=c\mod m\iff n^2-n-2c=0\mod 2m\\ $$ Since $\left(\frac{1+4c}{2m}\right)=\left(\frac{1+4c}{2}\right)^m=\left(\frac{1}{2}\right)^m=1$, a solution could be given by $n=\frac{1\pm\sqrt{1+4c}}{2}\mod 2m$ because $\sqrt{1+4c}$ must be odd. So I would attempt to do some generalization.
Case 1. Suppose $f(x)\in\mathbb{Q}[x]$ is some integer valued polynomial, i.e. $f(\mathbb{Z})\subseteq\mathbb{Z}$. Is it possible to estimate the image size $\#f(\mathbb{Z})/m\mathbb{Z}:=\#\{f(n)\mod m:n\in\mathbb{Z}\}$, algorithmically or by direct classification.
Case 2. Following case 1, if we restrict to where $f(x)\in\mathbb{Z}[x]$, could we estimate $$ \#f(\mathbb{Z})/m\mathbb{Z}=\#f(\mathbb{Z}/m\mathbb{Z})? $$ Specifically, if that equals $m$, then $f:\mathbb{Z}/m\mathbb{Z}\rightarrow\mathbb{Z}/m\mathbb{Z}$ is bijective.
Case 3. Following case 2, if we further restrict $\deg f$ to be small (e.g. quadratic), what can we get?
Case 4. Following case 2, if we further restrict $m$ be a prime or prime power, what can we get?
Is this even computationally feasible? That being said, there's no need to count over all possible points from $\mathbb{Z}/m\mathbb{Z}$ solve for $f(n)=c\mod m$. On the other hand, suppose $\#f(\mathbb{Z})/m\mathbb{Z}< \mathbb{Z}/m\mathbb{Z}$, could we say anything about their gap, i.e. $\mathbb{Z}/m\mathbb{Z}-\#f(\mathbb{Z})/m\mathbb{Z}$, perhaps asymptotically?