We have
$$n^8 + n^6 + n^4 + 4 = (n^4 - n^3 + n^2 - 2n + 2)(n^4 + n^3 + n^2 + 2n + 3) = f(n)g(n)$$
The first factor $f(n)$ satisfies
$$f(n) = n^4 - n^3 + n^2 - 2n + 2 = n^3(n-1) + (n-1)^2 + 1$$
and hence $f(n) \ge 2$ for all $n \ge 2$. The second factor $g(n) = n^4 + n^3 + n^2 + 2n + 3$ is strictly greater than 2 for all $n \ge 2$. This only leaves the case $n = 1$ as a potential candidate for a prime, and indeed $f(1)g(1) = 1 \cdot 7 = 7$ is prime.