Question:
What conditions are necessary on $a,b,c,d$ so that the Möbius transformation $w=\frac{az-b}{cz-d}$ has only one fixed point?
Attempt: We examine $$ z=\frac{az-b}{cz-d}$$ to find that any fixed point of a Möbius transformation must satisfy $$cz^2 + (d-a)z - b =0.$$ Hence, there are either one or two fixed points in $\mathbb{R}$ OR two fixed points in $\mathbb{C}$ which are all characterized by $$z = \frac{(a-d) \pm \sqrt{(d-a)^2+4bc}}{2c}.$$ Hence, there can only be a fixed point when the discriminant $(d-a)^2+4bc$ is zero. Further, if there is only one fixed point, it must be in $\mathbb{R}$.
Can someone with more knowledge than me comment on if my approach is right? Is there a better way to solve this question?
First, you've forgotten the possibility that $c=0$. Then you get a single fixed point when $a\neq d$.
The rest of your answer is correct, but an alternative is to note that $(d-a)^2+4bc = (a+d)^2 -4(ad-bc)$. That's interesting becuase $a+d$ is the trace and $ad-bc$ is the determinant of the matrix:$$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
Note, if you allow $\infty$ in your domain and range, then the case $c=0$ is different, then $a=d$ and $b\neq 0$. This is a stronger answer, because what it means is that the general solution ($c=0$ or $c\neq 0$) is of the form:$$A = U\begin{pmatrix}x&y\\0&x\end{pmatrix}U^{-1}$$ with some invertible matrix $U$, with $xy\neq 0$. That's because the zero value of the discriminant means that the eigenvalues of the matrix must be the same, and the matrix can't be a diagonal matrix or then the transformation is the identity.
This can be stated briefly as: $A$ is not diagonalizable over $\mathbb C$.