The conclusion clearly holds if $n$ is even because in this case $2|(n^4+4^n)$. If $n$ is odd, let $n=2m+1$. Then, $$n^4+4^n = n^4+4^{2m+1}=n^4+4\cdot (2^m)^4=(n^2 + 2\cdot(2^m)^2 + 2\cdot n\cdot 2^m)(n^2 +2\cdot (2^m)^2 - 2\cdot n\cdot 2^m)$$ The last step is by Sophie Germain identity (see %%HREF%%3863%%). Now, all we need to show is the smaller of these two factors is greater than $1$. $$n^2 + 2\cdot (2^m)^2 - 2\cdot n\cdot 2^m =n^2 + 2^{2m+1}-n\cdot 2^{m+1}=(n-2^m)^2+2^{2m} > 1$$