Ukrainian National Mathematical Olympiad

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