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, valid in some interval,  $$f(x)=\sum_0^{\infty}a_k\frac{x^k}{k!} $$ and that 

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

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

So let me answer your question first for $\mathbb{N}$ rather than $\mathbb{Z}.$ Note that the binomial coefficients can be written  $\binom{n}{k}=\frac{(n)_k}{k!}$ Where $(n)_k=n(n-1)(n-2)\cdots(n-k+1)$ is the falling factorial.

Any real function $f(n)$ defined on $\mathbb{N}$  (a.k.a. a sequence) has a unique expansion (valid on all the domain)  $$f(n)=\sum_0^{\infty}a_k\frac{(n)_k}{k!}.$$

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

Here $a_k=(\Delta^kf)\ (0)$ where $\Delta$ is the difference operator which sends
 the sequence 

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

to the sequence 

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


>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 $\mathbb{Z}.$

That there is such an expansion for $\mathbb{N}$ is also clear if we examine the sequences $\binom{n}{k}:$

$$\begin{array}{cccccccc}
  1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \cdots \\
  0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\cdots\\
  0 & 0 & 1 & 3 & 6 & 10 & 15 & 21\cdots \\
  0 & 0 & 0 & 1 & 4 & 10 & 20 & 35 \cdots\\
  0 & 0 & 0 & 0 & 1 & 5 & 15 & 35\cdots 
\end{array}$$
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.