z-transforms of difference equations are pretty easy. Any term that looks like A y(k+n), transforms to A z^n Y(z) in the z-domain. Any linear difference equation with constant coefficients (which is what you have) will be make up of terms that look like this.
Then, algebraically solve for Y(z). Then inverse transform back using lookup tables. You will find that your example is in most tables, however, you usually have to do a little bit of algebraic manipulation to make it look exactly like the function in the table.
For example if your function looks like F(z) and the table only has z F(z), then the effect of the extra multiplication by z just means a time delay by one sample to your actual answer. Also, if your function is F(z) but your table gives you C F(z), then rewrite your function as 1/C (C F(z)), and then you can use the table.