How can it be shown that a Möbius transformation can have at most two fixed points unless it is f(z) = z?

Let $z \to \frac{az+b}{cz+d}$ $(a,b,c,d \in \mathbb{C}$, and $ z \not = \frac{-d}{c}$) be the Mobius transformotation. A point $z_{0}$ if fixed if and only if $\frac{az_{0}+b}{cz_{0}+d}=z_{0}$,so $cz_{0}^{2}+(d-a)z_{0}-b=0$. This is an equation of degree at most $2$, hence has at most two solutions!

Hint

Mobius transform is a function of the form $f(z)=\frac{az+b}{cz+d}$. So for fixed points you need to solve $f(z)=z$. This results in (a possibly) second degree equation in $z$ and that will have (at most) two roots.