Prove that if $A - A^2 = I$ then $A$ has no real eigenvalues

Hint: Apply your equation to an eigenvector.

$A-A^2=I\to A^2-A=-I\to A^2-A+I=0$

Since A satisfies its own characteristic equation (Cayley-Hamilton Theorem), replace A with $\lambda$ and see what you get....

Hint: Is there a real number $\lambda$ with $\lambda-\lambda^2=1$?