It seems too easy for a TST,so it may be wrong.The answer is 2^(n-m).Each leter from alphabet $A$ pair $2$ letters from alphabet $B$.Now,we will biject $an$ to $bn$.Consider some element from $an$.Now,on all places where one letter appears,replace it with his $2$ paired letters. Lemma:Let $S$ be the set of all binary strings with even lenght $d$ such that they have an odd number of $1$-s.Then $|S|$=2^(d-1). Proof:This is trivial.Consider all elements expect the last one.Now,the last element is determined by the binary string before him of lenght $d-1$,so we have 2^(d-1) such strings. Now,applying this lemma in our bijection,we easy get the answer.

I somehow feels that this is similar with IMO 2008 Problem 5, and I also exponential generating functions will help a lot here.

This problem is almost trivialized by generating functions. Letting $ A(x) $ and $ B(x) $ be the generating functions for $ \frac{a_n}{n!} $ and $ \frac{b_n}{n!} $ respectively we easily find that $ A(x) = \left(\sum_{k = 1}^{\infty}\frac{x^{2k}}{(2k)!}\right)^m $ and $ B(x) = \left(\sum_{k = 1}^{\infty}\frac{x^{2k - 1}}{(2k - 1)!}\right)^{2m}. $ But $ \left(\sum_{k = 1}^{\infty}\frac{x^{2k - 1}}{(2k - 1)!}\right)^2 = \sum_{k = 1}^{\infty}\frac{x^{2k}}{(2k)!}\sum_{j = 1}^{k}\binom{2k}{2j - 1} = \frac{1}{2}\sum_{k = 1}^{\infty}\frac{(2x)^{2k}}{(2k)!}. $ This implies that $ B(x) = \frac{A(2x)}{2^m} $ which immediately implies that $ \frac{b_n}{a_n} = 2^{n - m}. $