Carefully determine all divisors of $f(x)$ where $$ f(x)=x^n\in F[x]$$ note that $F[x]$ is a Field
So, $$ \underbrace{x^0\mid x^n,\ x^1\mid x^n,\dots,\ x^n\mid x^n}_{n+1}$$ making $n+1$ divisors.
ify on Associates? $x^n$ has an associate $a x^n$ because $f(x)=a^{-1} (a x^n)$.
The associate does divide $x^n$ since F is a field and evry element including $a$ is a unit.
$$ ax^n \mid x^n \equiv \exists a^{-1}:x^n=1_R(x^n)=(a^{-1})ax^n$$