The suggestion from the comments to use Siegel's Theorem about integral points on algebraic curves is vast overkill. The by far easier Hilbert's Irreducibility Theorem is sufficiently strong here:

Suppose that $f(\mathbb Z)\subseteq g(\mathbb Z)$. Write $f(X)-g(Y)=A_1(X,Y)A_2(X,Y)\cdots A_r(X,Y)$ with polynomials $A_i(X,Y)\in\mathbb Q[X,Y]$ which are irreducible over $\mathbb Q$. By Hilbert's Irreducibility Theorem, there is an infinite set $H$ of integers such that $A_i(h,Y)$ is irreducible for each $h\in H$ and each $i$. On the other hand, for each integer $h$ there is an integer $u$ such that $f(h)=g(u)$. So $f(h)-g(Y)$ and therefore some $A_i(h,Y)$ has an integral root. Running through $h\in H$, we see that there is some $i$ such that $A_i(h,Y)$ has an integral root for infinitely many $h\in H$.

So for this index $i$ there are infinitely many $h\in H$ such that $A_i(h,Y)$ is irreducible *and* has a rational root. So $A_i(X,Y)$ has degree $1$ in $Y$.

We obtain $f(X)-g(Y)=A(X,Y)(b(X)-Yc(X))$ for $A\in\mathbb Q[X,Y]$ and $b,c\in\mathbb Q[X]$. Setting $Y=b(X)/c(X)$ yields $f(X)=g(b(X)/c(X))$. Looking at poles shows that $c(X)$ actually is a constant. So $f(X)=g(b(X))$ for a polynomial $b(X)$ with rational coefficients.