Similar matrices and field extensions

If the fields are infinite, there is an easy proof.

Let $F \subseteq K$ be a field extension with $F$ infinite. Let $A, B \in \mathcal{Mat}_n(F)$ be two square matrices that are similar over $K$. So there is a matrix $M \in \mathrm{GL}_n(K)$ such that $AM = MB$. We can write:
$$M = M_1 e_1 + \dots + M_r e_r,$$
with $M_i \in \mathcal{M}_n(F)$ and $\{ e_1, \dots, e_r \}$ is a $F$-linearly independent subset of $K$. So we have $A M_i = M_i B$ for every $i = 1,\dots, r$. Consider the polynomial
$$P(t_1, \dots, t_r) = \det( t_1 M_1 + \dots + t_r M_r) \in F[t_1, \dots, t_r ].$$
Since $\det M \neq 0$, $P(e_1, \dots, e_r) \neq 0$, hence $P$ is not the zero polynomial. Since $F$ is infinite, there exist $\lambda_1, \dots, \lambda_r \in F$ such that $P(\lambda_1, \dots, \lambda_r) \neq 0$. Picking $N = \lambda_1 M_1 + \dots + \lambda_r M_r$, we have $N \in \mathrm{GL}_n(F)$ and $A N = N B$.