Notice that over the binary field $K$, $(x^{15}+1)^8 = x^{120} + 1$ by repeated squarings. So if we find the irreducible factorization of $x^{15} + 1$, we have that of $x^{120} + 1$ as an immediate consequence.
Since $15 = 3\cdot 5$, we already know a couple of easy factorizations of $x^{15} + 1$, as the sum of third and fifth powers, respectively. Let's pick the first of these and refine it:
$$ x^{15} + 1 = (x^5 + 1)(x^{10} + x^5 + 1) \tag{1}$$
where the minus signs that normally appear are replaced without error by plus signs over $K$.
Now we further factor:
$$ x^5 + 1 = (x + 1)(x^4 + x^3 + x^2 + x + 1) \tag{2}$$
This earlier Question elicits all the irreducible polynomials in $K[x]$ up to degree 5, and the Answers there inform us that the second factor in (2) is indeed irreducible.
It remains to get irreducible factors of the second factor in (1). We know there must be an irreducible factor $x^2 + x + 1$ hiding in there, because of the overall factor $x^3 + 1$ we did not pursue, and:
$$ x^{10} + x^5 + 1 = (x^2 + x + 1)(x^8+x^7+x^5+x^4+x^3+x+1) \tag{3}$$
Finally the last of these can be factored into irreducibles:
$$ x^8+x^7+x^5+x^4+x^3+x+1 = (x^4+x^3+1)(x^4+x+1) \tag{4}$$
Piecing all of this together gives us that $x^{15} + 1$ is the product of five distinct irreducible factors over $K$. Raising it to the eighth power gives $x^{120} + 1$.
The distinct divisors of $x^{15} + 1$ correspond to subsets of its five distinct irreducible factors, with the empty set corresponding to $1$ and the total set corresponding to $x^{15} + 1$ itself. So, in all there are $2^5 = 32$ divisors of $x^{15} + 1$.
The distinct divisors of $x^{120} + 1$ can be counted similarly, except that instead of having just the two choices for each distinct irreducible (to include or not), we choose between zero and eight of them. Thus in all there are $9^5 = 59049$ divisors of $x^{120}+1$.
