Maximum N that will hold this true Find the largest positive integer $N$ such that
$$\sqrt{64 + 32^{403} + 4^{N+3}}$$
is an integer
Is $N = 1003$?

Note that with $N=2008$ we have $ (2^{N+3}+8)^2=4^{N+3}+2\cdot 8\cdot 2^{N+3}+64=4^{N+3}+2^{2015}+64=64+32^{403}+4^N,$ so we conjecture that the maximal value is $2008$.
If $2^{2015}+2^6+2^{2N+6}$ is a perfect square then also $\frac1{64}$ of it, i.e.  $2^{2009}+1+2^{2N}=m^2$ for some $m\in\mathbb N$. 
But if $N> 2008$, we have $$(2^N)^2=2^{2N}<2^{2009}+1+2^{2N}<1+2^{N+1}+2^{2N}=(2^N+1)^2$$
so that $2^N<m<2^N+1$, which is absurd.
