Find all integer numbers $k$,$l$,$m$,$n$ for which $3^k$ • $5^l$ + $7^m$ = $2^n$ 
Find all integer numbers $k$,$l$,$m$,$n$ for which $3^k\cdot  5^l+7^m=2^n$.

I tried to find a decent modulo and look at ending digits, but nothing more. Thank you for your responses. 
Obviously is $k$ = $l$ = $m$ = $0$ and $n$ = $1$ a solution.
 A: From Robert's list: there is an elementary method to find all solutions to, say,
$$ 2^n - 7^m = 15. $$ Evidently we need to solve the same problem when replacing $15$ by these $1,3,5,9,15, 25.$ I do not see any reason to expect a proof that $k+l \leq 2.$ 
For the six problems mentioned, see
Exponential Diophantine equation $7^y + 2 = 3^x$
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$. 
Finding solutions to the diophantine equation $7^a=3^b+100$ 
Gottfried Helms came up with a different algorithm that allowed him to solve problems where my numbers got too big.
A: I find the following values:
$$ \matrix{ k & l & m & n\cr0 & 0 & 0 & 1 \cr
0 & 0 & 1 & 3 \cr
0 & 2 & 1 & 5 \cr
1 & 0 & 0 & 2 \cr
1 & 1 & 0 & 4 \cr
1 & 1 & 2 & 6 \cr
2 & 0 & 1 & 4 \cr
}$$
These are the only ones with $0 \le k,l,m \le 100$.
A: let (1): $3^k 5^l+7^m=2^n$ (supposing that k and l are greater or equal to 1).
Studying the equation modulo 3, you see that n is even, $n=2n'$. Studying now the equation at modulo 5, you see too that $n-m = 4a$, for some $a$. Pass the 7^m to the right, factorize and observe that $2^{n'}-7^{2a-n'}$ and $2^{n'}+7^{2a-n'}$ are coprime. The rest its  easy and you can finish without much dificulty.
