I guess you won't be satisfied with the answer $f=0$ and $g=1$. :)

But the answer is no even if you assume that $f$ and $g$ are nonconstant. For example, consider $f(x)=2x^3$ and $g(x)=(x^3-2)^3$. If $f(a)=g(b)$ for some $a,b \in \mathbb{Q}(\zeta)$, then one finds that $2$ is a cube in $\mathbb{Q}(\zeta)$, which is a contradiction.

I guess you won't be satisfied with the answer $f=0$ and $g=1$. :)

But the answer is yes even if you assume that $f$ and $g$ are nonconstant. For example, consider $f(x)=2x^3$ and $g(x)=(x^3-2)^3$. If $f(a)=g(b)$ for some $a,b \in \mathbb{Q}(\zeta)$, then one finds that $2$ is a cube in $\mathbb{Q}(\zeta)$, which is a contradiction.