We can check that $2^{a}+2^{b}$ is never divisible by 7 , so we must have $c<7$. The binary representation of $2^{a}+2^{b}$ has at most two 1 's. Writing 0 !, 1 !, 2 !, $\ldots,$ 6 ! in binary, we can check that the only possibilities are c=2,3,4$, giving solutions$(0,0,2),(1,2,3),(2,1,3)$,$(3,4,4),(4,3,4)$.