There are two cases to consider: a is odd and a is even.

Case 1: a is even.

Then bot a-1 and a+1 are odd.  This then forces b=0.  Then the only potential for a solution is to find a pair of powers of 3 whose difference equals 2.  1=3^0 and 3=3^1 is the only such pair giving rise to (a,b,c)=(2,0,1).

Case 2: a is odd.

Then a-1 and a+1 are both even.  Divide both sides by 4.  Then [(a-1)/2] * [(a+1)/2] = 2^(b-2) * 3^c.

[(a-1)/2] and [(a+1)/2] are consecutive integers.  Then a solution exists for every pair consecutive integers that consist of a power of two and a power of three.
The only known pairs are (1,2), (2,3), (3,4), and (8,9).  These give rise to solutions (a,b,c)=(3,3,0), (5,3,1), (7,4,1), (17,5,2) respectively.