Ukrainian National Mathematical Olympiad

The powers of $2$ give remainders $2$, $4$ and $1$ in division by $7$ with period $3$, in particular, $2^k\equiv 4(\mathrm{mod}7)$ if and only if $k\equiv 2(\mathrm{mod}3)$. But this means that $m^2\equiv 2(\mathrm{mod}3)$, which is impossible. This contradiction completes the proof.