Let $p$ be prime, and let $m,n \in \mathbb{Z}$ be such that $m \mid pn$ but $m \nmid n$.
Let $N^{pn}_n$ denote the norm function mapping $\mathbb{F}_{q^{pn}} \rightarrow \mathbb{F}_{q^n}$, defined by $$N^{pn}_n(x) = x\cdot x^{q^n} \cdot x^{q^{2n}} \cdots \cdot x^{q^{(p-1)n}}.$$
Let $\alpha \in \mathbb{F}_{q^n}^*$, and let $A = \{N^{pn}_n(x) : x \in \mathbb{F}_{q^m}^*\}$. Let $\alpha A$ denote the coset $\{\alpha N^{pn}_n(x) : x \in \mathbb{F}_{q^m}^*\}$.
Question: For $k<n$, $k \mid n$, when is $\alpha A \cap \mathbb{F}_{q^k}^*$ empty/non-empty?
Observations so far:
$\bullet$ The mapping $x \mapsto N^{pn}_n(x)$ maps $\mathbb{F}_{q^m}^*$ into $\mathbb{F}_{q^n}^*$. The image of $\mathbb{F}_{q^m}^*$ under this mapping has size $\frac{q^m-1}{\gcd(q^m-1,(q^{pn}-1)/(q^n-1))}$.
$\bullet$ The set $\alpha A \cap \mathbb{F}_{q^k}^*$ is non-empty if and only if there exists an element $\beta \in \mathbb{F}_{q^m}^*$ such that $(\alpha\beta^{(q^{pn}-1)/(q^n-1)})^{q^k-1} = 1$. (Note that $\beta^{(q^{pn}-1)/(q^n-1)}$ is just $N^{pn}_n(\beta)$.)
Regarding the last observation, I've been able to formulate a condition that is sufficient for $\alpha A \cap \mathbb{F}_{q^k}^*$ to be non-empty, namely that $\gcd(1+q^m+\cdots+q^{pn-m}, 1+q^k+\cdots+q^{m-k}) = 1$ (whatever that simplifies to). I got there by writing everything as a power of a primitive element of $\mathbb{F}_{q^{pn}}$ and then looking at the resulting linear congruences in the exponents.
However, this approach feels clumsy can caveman-like, I was hoping that there's a "nicer" way to go at this.
Edit: I've asked a more general (and perhaps cleaner) version here: Maximal proper subfields and the behavior of the norm.