It is a nice exercise to prove that the only solutions (positive integers $x$) of the equation on the title are products of Mersenne primes; with all exponents equal to $1$.
((see also: A046528 in the OEIS))
Question: It is true that the only solutions $A \in GF(2)[x]$ to the equation
\sigma(A) = x^a(x+1)^b
are products of distinct
Mersenne irreducible polynomials $M$ where this means
M = x^c(x+1)^d+1
and $M \in GF(2)[x]$ is irreducible.
Trivial example: $$ \sigma(x^2+x+1)=x(x+1). $$ As usual $\sigma(n)$ is the sum of all positive divisors of the positive integer $n$ and $\sigma(A)$ is the sum of all divisors (including $1$ and $A$) of the polynomial $A$ in $GF(2)[x])$.