Let $K$ a field, $P \in K[X]$ irreducible of degree $n$, $L$ an extension field of $K$ with degree $m$ and $d=\gcd(m,n)$. I want to show that for every $Q \in L[X]$ irreducible factor of $P$, $\frac{n}{d}$ divides the degree of $Q$ and that $P$ has at most $d$ irreducible factors in $L[X]$.
If $d=1$, the result is easy to prove, but I struggle to extend my proof when $m$ and $n$ are not coprime. Here is my unsuccessful attempt.
Let $\alpha$ be a specific root of $P$, such as $K[\alpha]=K[X]/P$. We consider $Q$ the irreducible factor of $P$ over $L$ which contains $\alpha$. Let $\delta$ be the degree $\alpha$ over $L$. We obtain the following graph, where $k \in \mathbb{N}$.
$$\begin{array}{ccccc} & & L(\alpha) \\ &\nearrow \delta & & \nwarrow k \\ L & & & & K(\alpha) \\ &\nwarrow m & & \nearrow n \\ & & K \end{array}$$
I tried to find an arithmetical argument, to prove that $\frac{n}{d}$ divides the degree of $Q$, but it seems that I need another algebraic argument to come to the conclusion.
In fact, you have to pick $\alpha$ a root of $Q$. Then $[L(\alpha):L]=\deg Q$. Moreover, since $Q\mid P$ we have $P(\alpha)=0$, so $P$ is the minimal polynomial of $\alpha$ over $K$. Then $[K(\alpha):K]=n$. We obtain $[L(\alpha):K]=m\deg Q=n[L(\alpha):K(\alpha)]$, so $n\mid m\deg Q$ and therefore $\frac nd\mid \deg Q$. This leads immediately to the conclusion of your question: $P$ has at most $d$ irreducible factors in $L[X]$.