Is there an obvious reason why $4^n+n^4$ cannot be prime for $n\ge 2$? I searched a prime of the form $4^n+n^4$ with $n\ge 2$ and did not find one with
 $n\le 12\ 000$. 


*

*If $n$ is even, then $4^n+n^4$ is even, so it cannot be prime.

*If $n$ is odd and not divisible by $5$ , then $4^n+n^4\equiv (-1)+1\equiv 0 \pmod 5$.
So, $n$ must have the form $10k+5$.
For $n=35$ and $n=55$, the number $4^n+n^4$ splits into two primes
with almost the same size.
So, is there an obvious reason (like algebraic factors) that there is no prime I am looking for ?

If $n$ is even, so will be $4^n+n^4$ and the later will definitely be $>2$ and hence composite 
Else
$$4^n+n^4=(2^n)^2+(n^2)^2=(n^2+2^n)^2-2\cdot2^n\cdot n^2$$
$$=(n^2+2^n)^2-(n2^{\frac{n+1}2})^2$$
As $n$ is odd $\iff n+1$ is even $\implies\dfrac{n+1}2$ is an integer
$$4^n+n^4=(n^2+2^n+n2^{\frac{n+1}2})(n^2+2^n-n2^{\frac{n+1}2}) $$
Establish that both factors are $>1$
