We will prove that for all positive integers of the form $n = 3p$ (where $p > 3$ is a prime number)
$$n \mid 2^n - 8.$$
By Fermat's little theorem we have $2^p \equiv 2 \pmod{p}$, so it follows that
$$2^{3p} - 8 = (2^p)^3 - 8 \equiv 2^3 - 8 = 0 \pmod{p}. \quad (9)$$
Analogously, since $3p$ is an odd number, we have
$$2^{3p} - 8 \equiv (-1)^{3p} - 2 \equiv -3 \equiv 0 \pmod{3}. \quad (10)$$
Since $3$ and $p$ are relatively prime, from (9) and (10) we conclude that
$$2^{3p} - 8 \equiv 0 \pmod{3p}.$$
There are infinitely many prime numbers greater than $3$, hence the assertion follows.