Do there exist any odd prime powers that can be represented as $n^4+4^n$? 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.
 A: I think that your argument works as long as the smaller of your two factors
$$n^2+2^n−2^{m+1}n$$
is not equal to $1$. In the case of $n=1$, it does equal $1$, and you get Henry's counterexample. For larger $n$,
$$\begin{align}n^2+2^n−2^{m+1}n&=n^2+2^n-2^{(n+1)/2}n\\
&=n^2+\sqrt{2^n}(\sqrt{2^n}-n\sqrt{2})\end{align}$$
will be larger than $1$ for sure once $n$ is large enough for $\sqrt{2^n}-n\sqrt{2}>0$. This is equivalent to $2^n>2n^2$. This inequality is true for $n=7$. Calculus can confirm that it holds once $n$ is beyond $7$ too:
Let $f(n)=2^n$ and $g(n)=2n^2$. Then $f''(n)=2^n(\ln(2)^2)$ and $g''(n)=4$. So certainly for $n\ge7$, $f$ has the larger second derivative. The first derivative of $f$ at $n=7$ is $128\ln(2)$, which is larger than $g$'s first derivative at $n=7$, which is $28$. So for all $n\ge7$, $f$ has the larger first derivative. The same argument applies to $f'$ and $g'$, and so we conclude that $f(n)>g(n)$ when $n\ge7$.
We can directly check that $n^2+2^n-2^{(n+1)/2}n\neq1$ when $n$ is $3$ or $5$.
A: For odd $n>1$ the expression is never prime, it can be factored into a product where each factor is the sum of two squares:
$$[((2^m)+n)^2 +(2^m)^2][(2^m)^2 +((2^m)^2-n)^2],$$
where $m = (n-1)/2$.
Looks like I didn't read the question properly thanks for the feedback.
I will try and improve my answer. 
Let the first factor be $A$ and the second $B$. Then we have $AB = p^k$
with $k >1$, as if $k= 1$ then we have the case $AB =5$.
This means in $A$ and $B$, $n >1$, and so $m$ is at least $1$, (as $n$ is odd). 
Then it's easy to see that neither $A$ or $B$ is trivially $1$. So we can set $k = r+s$ and as $A > B$ we can put $A = p^r$, and $B = p^s$, with $r>s>0$. 
Now forming the difference of $A$ and $B$, means, we have;
$$A - B =  p^r - p^s = n 2^{m+2}. $$
But as $p>s>0$, we have $p$ divides $n$ ($p$ is an odd prime) but this means $p$ divides $4$, as we have assumed  $p^k = n^4 + 4^n$, with $k >1$, which is absurd. 
Therefore the original expression is only only the power of an odd prime for the case $p = 5$. 
A: There is the small counterexample 
$$5^1=1^4+4^1$$
A: For odd $n>1$ the expression is never prime, it can be factored into a product where each factor is the sum of two squares. 
$$[((2^m)+n)^2 +(2^m)^2][(2^m)^2 +((2^m)^2-n)^2]$$
where $m = (n-1)/2$
