Assuming N=Z+: (a+1)(a−1)=2b3c and gcd, so two cases:
1)\ \ a+1=2\cdot 3^c. Then 2^{b-1}+2=2\cdot 3^c, so 2^{b-2}+1=3^c. Clearly b-2\ge 1. If b-2=1, then (a,b,c)=(5,3,1). If b-2\ge 2, then 1\equiv (-1)^c\pmod{4}, so c=2k. Then 2^{b-2}=\left(3^k+1\right)\left(3^k-1\right), so 3^k+1, 3^k-1 are powers of 2, so 3^k=3, so k=1, so (a,b,c)=(17,5,2).
2)\ \ a-1=2\cdot 3^c. Then 2\cdot 3^c+2=2^{b-1}, so 3^c+1=2^{b-2}, so 1\equiv (-1)^{b-2}\pmod{3}, so b-2=2k, so 3^c=\left(2^k+1\right)\left(2^k-1\right), so 2^k+1,2^k-1 are powers of 3, so 2^k=2, so (a,b,c)=(7,4,1).