Notice that $$\frac{1}{2}ab\sin \angle C=S=\frac{1}{4}(c^2-a^2-b^2)=\frac{1}{4}(-2ab\cos \angle C)=-\frac{1}{2}ab\cos \angle C$$and therefore $$\sin \angle C = -\cos \angle C \Rightarrow \tan \angle C =-1.$$However, $135^{\circ}$ is the only angle between $0$ and $180$ degrees that has a tangent of $-1$, so we are done.