In a finite field $GF(2^n)$, $u^2+u=0$ has only two solutions: $u=0, u=1$. (Not sure I can prove why this is true, probably something to do with invertibility.)
In a quotient ring $GF(2)[x]/p(x)$ where $p(x)$ is reducible, $u^2+u=0$ can have other solutions. For example with $p(x) = x^9+x^8+x^7+x^3+1=(x^7+x+1)(x^2+x+1)$, there is a solution $u = x^7+x+1$.
Is there a technique to find all solutions of this type, and anything special about the values? (presumably this is equivalent to finding polynomials $u,v$ in the polynomial ring $GF(2)[x]$ such that $u^2+u+vp(x) = 0$)
In the characteristic 2 case, essentially your question is equivalent to finding all idempotents of the ring $R = GF(2)[x]/p(x)$. If $p(x)$ is irreducible, $R$ is a finite field and there are only two trivial solutions, as you stated.
Suppose $p(x) = p_1(x)p_2(x) \cdots p_k(x)$ splits into $k$ distinct irreducible factors, there will be $2^k$ idempotents, including the trivial ones. Indeed, $f(x)$ is idempotent if $f(x) \equiv \epsilon_s \pmod{p_s(x)}$, where $\epsilon_s = 0$ or $1$ for $s = 1, \cdots, k$. For a fixed set of $(\epsilon_s)_{s = 1}^k$ there is a unique $f(x) \in R$ by CRT.
Generalization to a non square-free $p(x)$ should be quite doable.