Number of ordered Pairs satisfying $4^m-3^n=1$ Find the Number of ordered Pairs $(m,n)$ of positive integers satisfying $4^m-3^n=1$
Mt try:
Trivially $m=n=1$ satisfies
Let $m \gt 1$
$$4^m-3^n=(1+3)^m-3^n=1$$
$\implies$
$$3\binom{m}{1}+3^2\binom{m}{2}+3^3\binom{m}{3}+\cdots+3^m=3^n$$
Now since LHS is not a power of $3$ and RHS is, this is possible only when $m=1$
Hence the only ordered pair is $(1,1)$
is this the right way?

Let $(m,n)$ be a pair with $n>1$ such that $4^m-3^n=1$. Looking at both sides modulo $4$, we see that $n$ must be odd, so of the form $n=2k+1$. Thus we now have $4^m-3\cdot 9^k=1$.
Now looking at the equation modulo $9$, we see that $m$ must be a multiple of $3$, so of the form $m=3l$. Thus we have $64^l-3\cdot 9^k=1$.
However, comparing both sides modulo $7$, we now must have $1-3\cdot 2^k\equiv 1\pmod 7$, or $2^k\equiv 0\pmod 7$. This is clearly impossible, hence $(1,1)$ in fact is the only solution.
PS: I use $n>1$, or, equivalently, $k>0$, to say $3\cdot 9^k\equiv0\pmod9$.
