 Solution

The Gauss map sends $p$ to the chosen unit normal $N(p)$. Differentiating $⟨N,N⟩=1$ shows that $d{N}_p$ maps $T_{p}S$ into itself. The shape operator $S_p=-d{N}_p$ is self-adjoint because its associated second fundamental form is symmetric. Its eigenvalues $k_1,k_2$ are the principal curvatures. A point is umbilical when $k_1=k_2$, and the surface is minimal when its mean curvature $H=(k_1+k_2)/2$ vanishes everywhere.

In an orthonormal principal basis, $d{N}_p$ has diagonal matrix $\mathrm{diag}(-k_1,-k_2)$. It is conformal exactly whenwhich is equivalent to $k_1^2=k_2^2\ne 0$. At a nonumbilical point this says $k_1=-k_2$, exactly the condition $H=0$. Hence, when there are no umbilical points, the surface is minimal if and only if its Gauss map is conformal.