Proof
(related to Proposition: Characterization of Dependent Absolute Values)
 By hypothesis, $(F,+,\cdot)$ is a field with two absolute values $\cdot_1$ and $\cdot_2$ defined on it.
 Assume, $\cdot_1$ and $\cdot_2$ are dependent.
 Then for $x\in F$ we have $x_1 < 1\Longleftrightarrow x_2 < 1.$
 Let $x_0\in F$ with $x_0_1 > 1,$ then $x_0_2 > 1.$
 For an arbitrary $x\in F$ with $x\neq 0$, we have for some positive real number $\lambda > 0$ $$x_1=x_0_1^\lambda.$$
 For all real numbers $\epsilon > 0$ we get the approximations $$x_1^{\lambda\epsilon} < x_0_1^\lambda< x_1^{\lambda+\epsilon},$$ implying $$\left\frac{x^{\lambda\epsilon}}{x_0^\lambda}\right_1 < 1 < \left\frac{x^{\lambda+\epsilon}}{x_0^\lambda}\right_1.$$
 Since $\cdot_1$ and $\cdot_2$ are dependent, we get $$\left\frac{x^{\lambda\epsilon}}{x_0^\lambda}\right_2 < 1 < \left\frac{x^{\lambda+\epsilon}}{x_0^\lambda}\right_2,$$ implying $$x_2^{\lambda\epsilon} < x_0_2^\lambda< x_2^{\lambda+\epsilon}$$ for all $\epsilon > 0,$ implying
$$x_2^{\lambda} = x_0_2^\lambda.$$
 Obviously, $\lambda$ exists and is a positive real number.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

References
Bibliography
 Lang, Serge: "Algebra  Graduate Texts in Mathematics", Springer, 2002, 3rd Edition