Let $k = \mathbb{F}_{p^n} = \mathbb{F}_q$ finite field of $q = p^n$ and $[K:k]=2$ Galois extension of degree 2. Then $K = \mathbb{F}_{q^2} = \mathbb{F}_{(p^n)^2} = \mathbb{F}_{p^{2n}}$. It is generated by $\sigma : x \mapsto x^q$ the Frobenius automorphism. I want to calculate how many elements there are in $K^\times = K - \{ 0 \}$ such that $N_{K/k}(x) = \sigma(x)x = 1$.
As a numerical example let $p=2$ and $n=10$, then $k = \mathbb{F}_{2^{10}}$ and $K = \mathbb{F}_{2^{20}}$.
On
A different approach is given by the following argument (not using Hilbert 90). Pick an element $\gamma\in K\setminus k$. Let $z$ be an arbitrary element of $k$. Consider $$ x=\frac{z+\gamma}{z+\gamma^q}\in K. $$ We have $$ \sigma(x)=\frac{\sigma(z)+\sigma(\gamma)}{\sigma(z)+\sigma(\gamma^q)}=\frac{z+\gamma^q}{z+\gamma}=\frac1x. $$ Therefore $N(x)=1$. It is easy to show that different choices of $z$ yield different elements $x$, so there are at least $q$ elements in $\mathrm{ker} N_{K/k}$. Furthermore, obviously $1\in \mathrm{ker} N_{K/k}$ and $x=x(z)\neq1$ for all $z\in k$. Therefore $|\mathrm{ker} N_{K/k}|\ge q+1$. On the other hand $$ N_{K/k}(x)=x^{q+1}, $$ so there cannot be more than $q+1$ solutions to the equation $N_{K/k}(x)=1$.
Note that the construction is a cousin of the parametrization of points on the complex unit circle (other than $z=1$) by the recipe $$ z=\frac{x+i}{x-i},\quad x\in\mathbb{R}. $$
Admittedly Hilbert 90 also has a lot of appeal here.
By Hilbert 90, the kernel of $N_{K/k}: K^\times \to k^\times$ is the image of the homomorphism $\phi: K^\times \to K^\times, x \mapsto \sigma(x)/x$. The kernel of $\phi$ is $\{x \in K^\times : \sigma(x) = x \} = k^\times$, so we have an exact sequence $$1 \longrightarrow k^\times \longrightarrow K^\times \overset{\phi}{\longrightarrow} \operatorname{ker}N_{K/k} \longrightarrow 1.$$ Therefore, $$|\operatorname{ker}N_{K/k}| = \frac{|K^\times|}{|k^\times|} = \frac{q^2-1}{q-1} = q+1.$$