This is the same answer as given several times already. It is prefaced with a few facts on Taylor series to make it seem familiar. First recall that a real function $f(x)$ well enough behaved at $x=0$ has a Taylor series $$f(x)=\sum_0^{\infty}a_k\frac{x^k}{k!} $$ valid in some interval. And

f(x) is a polynomial if there is an $K$ with $a_k=0$ when $k \gt K$ (and the interval of validity is the entire real line.)

Similarly, any real function $f(n)$ defined on $\mathbb{N}$ (a.k.a. a sequence) has a unique expansion $$f(n)=\sum_0^{\infty}a_k\frac{(n)_k}{k!}.$$

valid on all the domain. And

$f$ is a polynomial (on $\mathbb{N}$) exactly if there is an $K$ with $a_k=0$ when $k \gt K.$

This leaves several things to explain.

For the Taylor series, $a_k=(D^kf)\ (0)$ where $D=\frac{d}{dx}$ is the differential operator which sends $f(x)$ to the derivative $f'(x)$.

For sequences on $\mathbb{N}$, $a_k=(\Delta^kf)\ (0)$ where $\Delta$ is the difference operator which sends the sequence

$f(0),f(1),f(2),f(3),\cdots$

to the sequence of differences

$f(1)-f(0),f(2)-f(1),f(3)-f(2),f(4)-f(3)\cdots$

Also, $(n)_k=n(n-1)(n-2)\cdots(n-k+1)$ is the falling factorial.
Of course the question concerned $\mathbb{Z}$ rather than $\mathbb{N}.$

For the domain $\mathbb{Z},$ we can first restrict to $\mathbb{N}$, determine if we have a polynomial and, if so, then check that the expansion is valid on all of $\mathbb{Z}.$

The expressions $\frac{(n)_k}{k!}$ were used point out the relation to Taylor series. Note, however, that $\frac{(n)_k}{k!}=\binom{n}{k}$ is a binomial coefficient.

Examination of these sequences $\binom{n}{k}$ make it clear that they constitute a basis (of sorts) for the space of sequence defined on $\mathbb{N}.$ The first few are

$$\begin{array}{cccccccc}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \cdots \\
0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\cdots\\
0 & 0 & 1 & 3 & 6 & 10 & 15 & 21\cdots \\
0 & 0 & 0 & 1 & 4 & 10 & 20 & 35 \cdots\\
0 & 0 & 0 & 0 & 1 & 5 & 15 & 35\cdots
\end{array}$$
Clearly there is a unique linear combination of these sequences for any target sequence. Although the sum is potentially infinite, it is finite for each fixed value of $n.$

Furthermore, $f$ takes integer values on all of $\mathbb{N}$ exactly if the $a_k$ are all integers.