Solution:
The number $m = 8^{n} + 47$ is never prime.

If $n$ is even, say $n = 2k$, then $m = 64^{k} + 47 \equiv 1 + 2 \equiv 0 \pmod{3}$. Since also $m > 3$, then $m$ is not prime.

If $n \equiv 1 \pmod{4}$, say $n = 4k + 1$, then $m = 8 \cdot (8^{k})^{4} + 47 \equiv 3 + 2 \equiv 0 \pmod{5}$. Since also $m > 3$, then $m$ is not prime.

If $n \equiv 3 \pmod{4}$, say $n = 4k + 3$, then $m = 8(64^{2k+1} + 1) \equiv 8((-1)^{2k+1} + 1) \equiv 0 \pmod{13}$. Since also $m > 13$, then $m$ is not prime.