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Solve the following equation in natural numbers:
$3^m-7^n=2$.

Source: AOPS

I think n=2 and m=1 . Am I correct?

Proof please.

There are no smaller solutions than $(m,n)=(2,1)$ , when $3^2 - 7^1 = 2$, so let's look for larger ones.
Write the equation as $3^2(3^{m-2} - 1) = 7(7^{n-1}-1)$.
We have $3^2 \mid 7^{n-1}-1$; the multiplicative order of $7$ modulo $3^2$ is $3$, so we need $3\mid n-1$ and then $19\mid 7^3-1 \mid 7^{n-1}-1$.
We have $19 \mid 3^{m-2}-1$; the multiplicative order of $3$ modulo $19$ is $18$, so we need $3\mid n-1$ and then $37\mid 3^{18}-1 \mid 3^{m-2}-1$.
We have $37 \mid 7^{n-1}-1$; the multiplicative order of $7$ modulo $37$ is $9$, so we need $9\mid n-1$ and then $3^3\mid 7^9-1 \mid 7^{n-1}-1$.
But $3^3 \mid 3^2(3^{m-2} - 1)$ implies $3 \mid 3^{m-2} - 1$, impossible. Thus there are no other solutions.