Solving for impulse response
Your characteristic polynomial is indeed

$$\lambda^2-4\lambda+4=(\lambda-2)^2$$

If you had two distinct roots $r_1$ and $r_2$ the general solution would have the form you suggested:

$$y(n)=C_1r_1^n+C_2r_2^n$$

However, in your case you have a double root $r=2$ which results in a general solution of the following form:

$$y(n)=C_12^n+C_2n2^n\tag{1}$$

The constants $C_1$ and $C_2$ are chosen to match the initial conditions and the input signal $x(n)$, which in order to compute the impulse response is a delta impulse $\delta(n)$. Assuming $y(-1)=y(-2)=0$ we get

$$y(0)=x(0)-x(-1)=1\\
y(1)=4y(0)+x(1)-x(0)=4+0-1=3$$

Comparing to (1) gives

$$y(0)=C_1=1\\
y(1)=2C_1+2C_2=3\Rightarrow C_2=1/2$$