The answer is negative. First notice that the sequences are subsets of the positive integers. Now notice that $a_n > 14a_{n-1} > 14^2a_{n-2} > \dots > 14^{n-1} a_1 = 14^n$ While $b_n < 6b_{n-1} < \dots < 6^{n-1}b_1 = 6^{n-1}.14$ But notice that $14^{n} > 6^{n-1}.14$, for every integer greater than $2$. Meaning that $a_n > 14^n > 6^{n-1}.14 > b_n$. Meaning that we won't ever have that $a_n=b_n$, for some $n \geq 2$.

@above,you may misunderstand the question,not to ignore the possiblity $a_{m}=b_{n}$(m,n are different)

bumpbump

bump again.

The answer is yes. We will prove that for any $k \geq 0$, $a_{2k+1} = b_{3k+1}$ holds. It is easy to prove by recurrence relations that if $\alpha = \sqrt{2} + 1$ and $\beta = \sqrt{2} - 1$, then $a_n = \alpha^{3n} + (-1)^n\beta^{3n}$ and $b_n = \alpha^{2n+1} - \beta^{2n+1}$. Now it is clear that $a_{2k+1} = b_{3k+1}$ by trivial algebra. Thus, there are infinitely many numbers occuring in both sequences.