Number theory divisibility check question

Question

N = $2^{744} - 1$. Prove N is divisible by $2^{93}+2^{47}+1$. I have no idea how to proceed. (edit: removed first part as I got the answer)

Answer 1

If $2^{31/2}=a,N=a^{48}-1$

$d=a^6+\sqrt2a^3+1$

$d$ will divide

$(a^6+1)^2-(\sqrt2a^3)^2=a^{12}+1$

which again divides $N$

Answer 2

By Aurifeuillean_factorization, $ 2^{186}+1=(2^{93}+2^{47}+1)(2^{93}-2^{47}+1),$

so $(2^{93}+2^{47}+1)$ divides $2^{186}+1.$

Then use $n+1$ divides $n^4-1=(n+1)(n-1)(n^2+1) $ with $n=2^{186}$ and you're done.