The problem statement is: if $\rho(z)=0$ with $z$ a double root, the corresponding difference equation has the solution $y_n = nz^n, \rho(E)y_n = 0, E^my_n=y_{n+m}$.
Here's what I did: Start with the polynomial applied to $y_n$, where the polynomial can be generalized by the following product: $$\rho(E) = \prod_{i=1}^{i=N}(E-x_i)$$ As the problem statement indicates, $z$ is a double root, so that the polynomial can be written as: $$\rho(E) = (E-z)^2\prod_{i=1}^{i=N-2}(E-x_i)$$ Let $\tilde{\rho}(E) = \prod_{i=1}^{i=N-2}(E-x_i)$. Then, \begin{align*} \rho(E)y_n &= \tilde{\rho}(E)(E-z)^2y_n \\ &= \tilde{\rho}(E)(E^2-2Ez+z^2)y_n \\ &= \tilde{\rho}(E)(E^2y_n-2zEy_n+z^2y_n) \end{align*} Inserting $E^my_n = y_{n+m}$ gives the following: \begin{align*} \rho(E)y_n &= \tilde{\rho}(E)(y_{n+2}-2zy_{n+1}+z^2y_n) \end{align*} Finally, $\tilde{\rho}(E)$ cannot be zero, so it must be the other polynomial. This is where I'm not sure what to do. If I substitute $y_n = nz^n$ I get zero - but is this correct? Or should I solve the polynomial and find the solution to be $y_n$?