I don't know of a fast way to solve the decision version of discrete log. Your intuition that factoring would be useful was correct; there is a efficient reduction to factoring $n$.
Given $a^x=b \bmod n$ for known $a$,$b$ to tell if a solution exists:
(1) Factor the modulus into prime powers. By the Chinese remainder theorem a solution exists iff it exists mod each prime power
(2) For each of these factor $p$-1
(3) Compute the order of $a$ and $b$ mod $p^k$ using the factorization of $p$-1 (can be done quickly using a generalization of Euler's criterion)
(4) A solution exists iff the order of $b$ divides the order of $a$.
(1) It is unknown if factoring $n$ reduces to being able to solve the decision discrete log mod $n$.
(2) Its not hard to show factoring $n$ reduces to the computational version of discrete log mod $n$ (discrete log can be used to compute the order of any element mod $n$ and there is a well known way to factor $n$ given this. It's described on the Wikipedia page for Shor's algorithm).
(3) Given the factors of $n$, discrete log mod $n$ reduces to discrete log mod $p$. It is unknown if a fast method for factoring any number would imply that the discrete log mod primes can also be solved quickly.