Find all pairs of natural numbers (m,n) satisfying $2^{m}-3^{n}=1$
My attempt: $$2^{m}-3^{n}=1$$ $\implies 2^m\equiv 1 \pmod3$ $\implies m=2k$ for some integer $k$. Hence, $$((2^{k})^{2}-1)=3^{n}$$ This implies, $$(2^k-1)(2^k+1)=3^n$$ But, $(2^k-1,2^k+1)=1$ , this implies that one among $2^k-1$ and $2^k+1$ is equal to $1$. It is obvious that $2^k-1<2^k+1$.Therefore, $2^k-1=1$ and $k=1$. This gives, $m=2$ and $n=1$.$(m,n)=(2,1)$ is the only solution of $2^{m}-3^{n}=1$.Does this look alright? Are there any other solutions besides $(2,1)$. Also, does my solution look neat(I don't think it does)?