= Solution
A minimal surface can have a planar point, where $k_1=k_2=0$. There $dN=0$, which fails the stipulated conformality condition $\lambda>0$. Thus minimality need not make the Gauss map conformal everywhere; a plane is the simplest counterexample.
Conversely, conformality gives $|k_1|=|k_2|\ne0$. It permits either $k_1=-k_2$, which is minimal, or $k_1=k_2\ne0$, which is umbilical and nonminimal. On a round <unit sphere>[sphere], the Gauss map is conformal while both principal curvatures are equal and nonzero. Thus conformality alone does not imply minimality.
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