Finding the $\mathcal Z$-transform of $((-\frac{1}{3})^n + \frac{1}{2})^n \mu[n-2]$?
Since $\left| -\frac{1}{3}\right|&lt; \frac{1}{2}$ and $n$ starts at $2$, one can think about an approximation based on the first terms of the polynomial development of $a_n = \left(\frac{1}{2}\right)^n\left(1+2\left(-\frac{1}{3}\right)^n\right)^n $. Using a binomial (polynomial) development up to degree $2$ on the second term only, one gets:

$$a_n \approx b_n= \left(\frac{1}{2}\right)^n + 2n\left(-\frac{1}{6}\right)^n + 2n(n-1)\left(\frac{1}{18}\right)^n \,.$$

Details: since coefficients are zero for $n&lt;2$, for $n=2$, 

$$a_2=\left(1+2\left(-\frac{1}{3}\right)^2\right)^2=  b_2 = 1+ {2 \choose 1}\left(2\left(-\frac{1}{3}\right)\right)+ {2 \choose 2}\left(2\left(-\frac{1}{3}\right)\right)^2$$

exactly. Then for $n&gt;2$, the term $c_n=2\left(-\frac{1}{3}\right)^n$ tends to be small. You can approximate $(1+c_n)^n$ by  $1+ {n \choose 1}c_n+ {n \choose 2}c_n^2$. 

The following graphs provide an evaluation of the approximation, up to $n=8$. Top graph superimposes $a_n$,  @MaximilianMatthé's approximation (MM), and what we get using the first (D0), the first two (D1), or the three terms (D2) of the above series. $D0$ (only $\left(\frac{1}{2}\right)^n$) is clearly insufficient. The others are quite close. The bottom graph shows the absolute approximation errors. As you can see, $D1$ and $D2$ converge very fast when $n$ grows, and are already quite good for the first terms. 



The nice thing with these approximations is that you not only reuse the classical properties of  the $\mathcal{Z}$-transform: linearity, time-shift and scaling, but also differentiation (up to order $2$), because of the factors $n$ and $n(n-1)$:

$$ n x_n \to -z\frac{dX(z)}{dz}\,.$$

A Matlab code for reproduction:

n = (2:8)';
s = ((-1/3).^n+1/2).^n;
s0 = ((-1/3).^n+(1/2).^n);
sa = ((1/2).^n);
sb = ((1/2).^n).*(1+2.*n.*(-1/3).^n);
sc = (1/2).^n+2.*n.*(-1/6).^n + 2.*n.*(n-1).*((1/18).^n);

subplot(2,1,1)
plot(n,[s,s0,sa,sb,sc],'.-');axis([0 max(n) -Inf Inf]); grid on
legend('Orig.','MM','D0','D1','D2')
h=xlabel('$n$ index');set(h,'INterpreter','latex');title('Series')
subplot(2,1,2)
semilogy(n,abs([nan(size(s)),s0-s,sa-s,sb-s,sc-s]),'.-');axis([0 max(n) -Inf Inf]); grid on
legend('Orig.','MM','D0','D1','D2')
h=xlabel('$n$ index');set(h,'INterpreter','latex');title('Absolute approximation errors')