In triangle $ABC,$
\[a^4 + b^4 + c^4 = 2c^2 (a^2 + b^2).\]Enter the possible values of $\angle C,$ in degrees, separated by commas.
From the Law of Cosines,
\[a^2 + b^2 - c^2 = 2ab \cos C.\]Squaring this equation, we get
\[a^4 + b^4 + c^4 + 2a^2 b^2 - 2a^2 c^2 - 2b^2 c^2 = 4a^2 b^2 \cos^2 C.\]From the given equation, $a^4 + b^4 + c^4 = 2a^2 c^2 + 2b^2 c^2,$ so
\[2a^2 b^2 = 4a^2 b^2 \cos^2 C.\]Then
\[\cos^2 C = \frac{1}{2}.\]Hence, $\cos C = \pm \frac{1}{\sqrt{2}}.$  Therefore, the possible values of $\angle C$ are $\boxed{45^\circ, 135^\circ}.$

If we take $a = \sqrt{2}$ and $b = c = 1,$ then $\angle C = 45^\circ.$  If we take $a = \sqrt{2},$ and $b = 1,$ and $c = \sqrt{5},$ then $\angle C = 135^\circ.$  Thus, both angles are achievable.