Hint $\ $ Write $\displaystyle\ f(x) = c_0 + c_1 {x \choose 1} + \cdots + c_4 {x \choose 4},\ $ possible since $\displaystyle {x\choose k}\,$ has degree $k$. Then
$$\begin{eqnarray} f(0) &=& c_0 \\ f(1) &=& c_0 + {1\choose 1} c_1 \\ f(2) &=& c_0 + {2\choose 1} c_1 + {2\choose 2} c_2 \\ f(3) &=& c_0 + {3\choose 1} c_1 + {3\choose 2} c_2 + {3\choose 3} c_3 \\ f(4) &=& c_0 + {4\choose 1}c_1 + {4\choose 2}c_2 + {4\choose 3} c_3 + {4\choose 4} c_4 \end{eqnarray}$$
Successively solving the above triangular system for $\,c_0,\ c_1,\ c_2,\ldots$ shows that all $\,c_i\in \Bbb Z$, since all $f(i), {j\choose k}\in \Bbb Z\,$ and each linear equation in $\,c_i\,$ is monic, i.e. has lead coef $= {i\choose i} = 1.\,$ Conversely, if all $c_i\in \Bbb Z$ then $f(n)\in \Bbb Z$ since all ${x \choose i}$ are integer valued.
Thus $f(x)$ is integer-valued $\iff$ its coefficients in a binomial basis are all integers.
Clearly the same proof works for polynomial of any degree. This is a well-known result of Polya and Ostrowski.