Assuming $\Bbb N=\Bbb Z^+$: $(a+1)(a-1)=2^b3^c$ and $\gcd(a+1,a-1)=2$, 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 $(a,b,c)=(7,4,1)$.