Set $h:=f-g$ then $h\not\equiv 0$ and ${{g(k)}\over {h(k)}}=n$ has an integer solution $k_n$ for every integer $n\ge 1$. Note that $|k_n|\rightarrow \infty$ as $n\rightarrow \infty$. Hence $\deg(h)\le 1$ and we can write ${g\over h}=px+q+{c\over {rx+s}}$ where $p,q,r,s,c\in Q$. There is a numerator $N$ such that $pZ+q \subset {1\over N}\cdot Z$ which is a discrete set $S$. Considering $pk+q+{c\over {rk+s}}=n$ for large $n$ we find $c=0$ since $pk+q, n\in S$. Hence $g,h$ have a common root $\alpha\in Q$ and then $f(\alpha)=g(\alpha)=0$.