TITLE: Find all integer solutions to $\displaystyle{2^k = 9^m+7^n}$
QUESTION [6 upvotes]: I can only find the trivial ones
$(m,n,k)$ as $(0,0,1), (1,1,4)$
Cannot find any more.  Are there any more values?  More imporantly how to show those are the only ones.

REPLY [5 votes]: We start like mercio, but then use a simple parity argument. By looking at the equation modulo $3$, we can see that $k$ must be even, say $k=2l$. Thus
$$7^n=(2^l-3^m)(2^l+3^m).$$
The two factors on the right are relatively prime, so
$$2^l-3^m=1 \qquad \text{and} \qquad 2^l+3^m=7^n,$$ 
and therefore
$$2^{l+1}=7^n+1 \qquad\qquad(\ast)$$
But $(\ast)$ has no solutions with $n>1$. Work modulo $16$.  Note that if $w$ is even, then $7^w \equiv 1\pmod{16}$, and therefore if $w$ is odd then $7^w \equiv 7 \pmod {16}$.  
If $n>1$ is even, then the right-hand side of $(\ast)$ is congruent to $2$ modulo $16$, while the left-hand side is divisible by $4$.  
If $n>1$ is odd, then the right-hand side of $(\ast)$ is congruent to $8$ modulo $16$. For the left-hand side to be large enough, it must be a large power of $2$, so cannot be congruent to $8$ modulo $16$.