Let $K$ be a finite field with $|K|=q$. Let $L|K$ be a finite field and $x\in L$, $x\neq 0$ be such that the order of $x$ in the multiplicative group $L^{\times}$ is $d$. I want to show that $[K(x):K]$ is the order of the residue of class of $q$ in $\mathbb Z_d^{\times}$.
We know that $[K(x):K]$ is equal to the order of the group Gal$(K(x)|K)$ which is a cyclic group generated by the Frobenius automorphism $f_q:K(x)\to K(x), x\mapsto x^q$. Suppose $[K(x):K]=m$. Thus Gal$(K(x)|K)=\{i,f_q,f_q^2,\ldots, f_q^{m-1}\}$. It follows that $x^{qm}=x\implies x^{qm-1}=1$. But order of $x$ in $L^{\times}$ is $d$. Thus $d|qm-1$. I could not proceed further. Any help is appreciated.
First of all note that it's $x^{q^m} = x$ and not $x^{qm} = x$. Indeed:
$$f^2(x) = f(x^q) = f(x)^q = (x^q)^q = x^{q^2}$$ $$f^3(x) = f(x^{q^2}) = f(x)^{q^2} = (x^q)^{q^2} = x^{q^3}$$ $$\cdots$$
Thus we have that $d \mid q^m - 1$ and so the order of $q$ in $\mathbb{Z}_d^{\times}$ (call it $n = \text{ord}_d(q)$) divides $m$. On the other side we have that $f^n(x) = x^{q^n} = x^{kd+1} = x$, for some $k \in \mathbb{Z}$, as $q^n \equiv 1 \pmod d$. Hence $f^n = 1$ in $\text{Gal}(K(x)/K)$. But the order of $f$ is $m$, so we have $m \mid n$.
Combining the two results we get that $m=n$ and so $[K(x):K] = \text{Gal}(K(x)/K) = m = n = \text{ord}_d(q)$