We need to have $2^m \equiv 8 (mod 9)$ and $2^m \equiv 6 (mod 10)$.
Note that $2^1 \equiv 2 (mod 9)$ and $2^7 \equiv 2 (mod 9)$ hence the cycle will repeat every 6 terms.
We have $2^2 \equiv 4 (mod 9) , 2^3 \equiv 8 (mod 9), 2^4 \equiv 7 (mod 9)$
$ 2^5 \equiv 5 (mod 9), 2^6 \equiv 1 (mod 9)$
Therefore, $m = 6k + 3$.
Also $ 2^1 \equiv 2 (mod 10), 2^2 \equiv 4 (mod 10), 2^3 \equiv 8 (mod 10), 2^4 \equiv 6 (mod 10)$
$2^5 \equiv 2 (mod 10)$ and the cycle continues.
Hence $4$ divides $m$. However, $m = 6k+3 = 3(2k+1)$ is an odd number. So contradiction.