I am having problems finding the mistake in my thought process here. One of the results from Ireland-Rosen on the topic:
If $F$ is a finite field with $q$ elements then for every $\alpha \in F^*$ , $x^n=\alpha$ has solutions iff $\alpha^{(q-1)/d} = 1$ where $d=\text{gcd}(n,q-1)$. If there are solutions, there are exactly $d$ solutions.
Now one of the suggested exercises is:
Let $F$ be a finite field with $q$ elements, $n\equiv 1 \pmod q$ and $K$ an extension of $F$ with $[K:F]=n$ then for every $\alpha \in F^*$, $x^n=\alpha$ has exactly $n$ solutions in $K$.
Don't these two results contradict each other? $K$ has $nq$ elements and therefore $x^n = \alpha$ has to have $\text{gcd}(nq-1,n)=(n(q-1)+n-1,n)=(n-1,n)=1$ or 0 solutions.
You're thinking of the number of elements computed from an index as if you were talking about a group. That's not how field index is defined.
The index, $[K:F]$ is the dimension of $K$ as a vector space over $F$. So if the index is $2$, $|K| = |F|^2$ and if the index is $n$, $|K| = |F|^n$.