Well, I wrote up a solution on it, but according to the place I found the problem, it isn't quite correct. Ah, I'm simply hoping someone will point out where I got wrong.
Now, let, $n^4+4^n = p^k$, where $p$ is an odd prime and $k$ is a positive integer.
Further, $p^k \equiv 1 \pmod 2$. Therefore, $n^4 + 4^n \equiv n^4 \equiv 1 \pmod 2$, and so $n \equiv 1 \pmod 2$. $n$ must be odd.
Okay, now let, $n = 2m +1$. Substituting it in $n^4+4^n$ and using the Sophie Germain inequality, we have,
$$n^4+4\cdot4^{2m} = n^4 + 4(2^m)^4 = (n^2 + 2^n+2^{m+1}\,n)(n^2 + 2^n-2^{m+1}\,n) = p^k$$
Now, as $p^k$ can only be factorized into smaller powers of $p$, let $n^2 + 2^n+2^{m+1}\,n = p^i$, and let $n^2 + 2^n-2^{m+1}\,n = p^j$ where $i+j= k$, obviously, and $i>j$.
Now consider this:
$$\begin{align} p^i - p^j & \equiv 0\\ 2\cdot2^{m+1}\,n = 2^{m+2}\,n &\equiv 0 \pmod p \end{align}$$
But, as $p$ is odd, $\gcd(p, 2) = 1$, so $n \equiv 0 \pmod p$.
Now look at this:
$$\begin{align} p^i + p^j &\equiv 0 \\ 2(n^2 + 2^n) &\equiv 0 \\ n^2 + 2^n &\equiv 0 \pmod p \end{align}$$
But we just established that $n \equiv 0 \pmod p$, so $2^n \equiv 0 \pmod p$.
Therefore, let $2^n = jp$ for some integer $j$.
Now, $2^n$ is its own prime factorization, which is unique according to the Fundamental Theorem of Arithmetic and does not include $p$. Therefore, the above statement is an impossibility! There exist no such $p$ and $n$, and no odd prime powers can be written as $n^4+4^n$.
Ah, well, that's it. Sorry for the tediousness of it. I've still no clue how to use $\LaTeX$.
Thank you everybody,
Cheers.
$p^i$
which looks like $p^i$ instead of p^i and everything looks much nicer ;) $\endgroup$