Let $m \geq 3$ be an odd integer. Prove that for any $a \in \mathbb{F}_{2^m}$, the equation $x^6+x+a=0$ has a zero or two distinct solutions in $\mathbb{F}_{2^m}$.
I have proven the following facts:
Let $a>1$ be an integer. For any positive $n,d\in \mathbb{Z}$, $\,$ $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. In particular, $\mathbb{F}_{p^d}\subseteq \mathbb{F}_{p^n}$.
Let $\phi$ denote the Frobenius map $x\mapsto x^p$ on the finite field $\mathbb{F}_{p^n}$. Then $\phi$ gives an isomorphism of $\mathbb{F}_{p^n}$ to itself. Also, $\phi^n$ is the identity map and no lower power of $\phi$ is the identity.
For any prime $p$ and any nonzero $a \in \mathbb{F}_p$ prove that $x^p-x+a$ is irreducible and separable over $\mathbb{F}_p$.
I don't know if any of these facts will help me prove this. I am currently studying Chapter 13 from Dummit and Foote. Any help on how to prove the statement is greatly appreciated, thank you.
***My question is not exactly the same as the post in Number of solns of $x^6+x=a$ in $\mathbb{F}_{2^m}$, where $m\geq 3$ is odd is same as number of solns of $x^2+ax+1=0$ because I found a counterexample for $x^2+ax+1=0$ having the same number of solutions as $x^6+ax+1=0$ in $\mathbb{F}_{2^m}$ whenever $m \geq 3$ is odd.