I want to find the solutions $(a,b)\in\mathbb{Z}^+\times\mathbb{Z}^+$ of $7^a+2=3^b$. One such solution is $(a,b)=(1,2)$.
Looking modulo $4$, we have $(-1)^a+2\equiv(-1)^b$, so $a$ and $b$ are of different parity.
Modulo $3$: $1+2\equiv 0$, which is always true.
Modulo $7$: $2\equiv 3^b$, so $b\equiv 2\pmod 6$. This means $a$ is odd.