Don't feel bad—in this particular proof, that particular formula is more than half the solution! This is a hard solution to come up with without having seen it (or something very similar to it) before.
I don't know if there's a name for this method, but here's how I would remember it: for any positive integer $a$, the polynomial $x^a-1$ certainly has $x=1$ as a root, and therefore $x^a-1$ is divisible by $x-1$. It turns out (and is easy to check) that the quotient of those two polynomials is very nice:
$$
x^a-1 = (x-1)(x^{a-1} + x^{a-2} + \cdots + x^2 + x + 1).
$$
In this solution, we're applying this factorization with $x=2^b$.
Indeed one can lift the above identity to one with two variables:
$$
x^a-y^a = (x-y)(x^{a-1} + x^{a-2}y + \cdots + x^2y^{a-3} + xy^{a-2} + y^{a-1}).
$$
This includes the familiar difference of squares $x^2-y^2=(x-y)(x+y)$ as the special case $a=2$. When $a$ is odd, there's also the similar identity
$$
x^a+y^a = (x+y)(x^{a-1} - x^{a-2}y + \cdots + x^2y^{a-3} - xy^{a-2} + y^{a-1}),
$$
where the signs in the second factor alternate. (This identity can be proved just by changing $y$ to $-y$ in the previous identity.)
The complete factorization of $x^a-1$ involves cyclotomic polynomials, so if you're looking for a word to attach to this factorization method, "cyclotomic" might be appropriate (although these particular identities barely scratch the surface of cyclotomic polynomials).
