Proving that $2^{744}-1$ is a multiple of $2^{248}+2^{124}+1$ and $2^{93}+2^{47}+1$ we have to prove that $2^{744}-1$ is multiple  of $2^{248}+2^{124}+1$ and $2^{93}+2^{47}+1$
I could prove first part as follows:
$$2^{744}-1=\left(2^{372}\right)^2 -1=(2^{372}-1) (2^{372}+1)$$
But $$2^{372}-1= \left(2^{124}\right)^3 -1$$ Using $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ we get another factor as $2^{248}+2^{124}+1$
But can i have any hint for second part

As I wrote in my comment, it is known that$$4a^4+1=4a^4+4a^2+1-4a^2=(2a^2+1)^2-(2a)^2=(2a^2+2a+1)(2a^2-2a+1)$$
You can try substituting $a$ with some number to show that $A=4a^4+1$ is multiple of $2^{93}+2^{47}+1$, and it won't be hard to prove that $2^{744}-1$ is multiple of $A$.
