Cute problem. Denote $a_1,a_2,b_1,$ and $b_2$ by $k,\ell, m,$ and $n$ respectively. Observe that $$\begin{cases} a_{j+2} = kF_j+\ell F_{j+1}\\ b_{j+2} = mF_j+nF_{j+1} \end{cases}$$where $F_i$ is the Fibonacci sequence defined by $F_1=F_2=1$ and $F_{n+2}=F_{n+1}+F_n$ for $n\geq 1$. This means that if $b_n/a_n$ is an integer infinitely often, there are infinitely many $j$ for which there exists a positive integer $c_j$ satisfying $\frac{mF_j+nF_{j+1}}{kF_j+\ell F_{j+1}}=c_j$, which in turn implies that $F_j(m-c_jk)=F_{j+1}(c_j\ell - n). $ By considering the sign of both sides of the above equation, we see that $c_j$ is bounded above, so by Pigeonhole, $c_j$ attains some fixed value $c$ infinitely often. Then for infinitely many $j$, $$F_j(m-ck)=F_{j+1}(c\ell - n).\qquad (\dagger)$$Now it's well-known that consecutive terms in the Fibonacci sequence are coprime, so $(\dagger)$ implies that $F_j\mid c\ell-n$. By taking $j$ large, we obtain $n=c\ell$, and similarly, by taking $j$ large in $F_{j+1}\mid m-ck$, we see that $m=ck$. Thus $b_1=ca_1$ and $b_2=ca_2$ for some fixed integer $c$, from which the desired result easily follows.