How to prove the cubic formula without root extraction

Your property is indeed true except possibly when $\gamma=0$, i.e. when $x^3+px+q$ has a double root.

If $\gamma\neq 0$, taking $\Delta=\frac{(mx^2-nx+2m^2)}{\gamma}$ you are done; notice the identity $$ (mx^2-nx+2m^2)^2=(n^2+m^3)(x^2+4m)+m(mx-2n)(x^3+px+q).$$

If $\gamma=0$ then $n=-\tau^3\neq 0$ by hypothesis, and $x^3+px+q$ can be rewritten as $x^3-3\tau^2x-2\tau^3=(x+\tau)^2(x-2\tau)$. If $x=2\tau$, your property is true with $\omega=1$. If $x=-\tau$ however, your property is true iff $F$ contains a primitive third root of unity. This is false when $F={\mathbb Q}$ and $\tau=1$ for example.