I first tried to factorize $z^3+z+2$ into the form $(z-a)(z-b)(z-c)$, where-after sum up the separate power series expansions of $\frac{A}{z-a} + \frac{B}{z-b} + \frac{C}{z-c}$, but I think there should be a simpler way.
I also tried using the binomial theorem, but this power expansion according to Wolfram Alfpa.
$$ \require{cancel} \frac{1}{z(z^2+1)+2} = \color{green}{\sum_{n=0}^{\infty} (-1)^n \cdot \bigg(\frac{1}{2}\bigg)^{n+1}z^n(z^2+1)^n} \\ = \sum_{n=0}^{\infty} (-1)^n \cdot \bigg(\frac{1}{2}\bigg)^{n+1} \cdot z^n \sum_{k=0}^{n} \binom nk z^{2\cancel{\color{red}{n}}k} \\ =\cancel{\sum_{n=0}^{\infty} (-1)^n \cdot \bigg(\frac{1}{2}\bigg)^{n+1} \cdot z^n \cdot 2^n \cdot z^{2n}} \\ \cancel{\sum_{n=0}^{\infty} (-1)^n \cdot \bigg(\frac{1}{2}\bigg) \cdot z^{3n}} \ . $$
$$\frac1{1+\dfrac{z+z^3}2}=1-\dfrac{z+z^3}2+\left(\dfrac{z+z^3}2\right)^2-\cdots$$ will yield terms in $z^0,z^1$ and $z^2$ which cannot be canceled by others, hence
$$\frac12\left(1-\frac z2+\frac{z^2}4\right).$$
By Taylor:
$$f(0)=\frac12$$
$$f'(x)=-\frac{3z^2+1}{(z^3+z+2)^2}\implies f'(0)=-\frac1{2^2}.$$
Then differentiating by sight, $$f''(0)=-\frac{6\cdot0}{\text{don't care}}-(-2)\frac{1\cdot1}{2^3}=\frac14.$$
Hence
$$\frac12-\frac z4+\frac{z^2}8.$$
A more systematic development is obtained with
$$f(x)(x^3+x+2)=1,\\ f'(x)(x^3+x+2)+f(x)(3x^2+1)=0,\\ f''(x)(x^3+x+2)+2f'(x)(3x^2+1)+f(x)6x=0 $$
then
$$2f(0)=1,\\ 2f'(0)+f(0)=0,\\ 2f''(0)+2f'(0)=0. $$
This can be continued at will and by the Leibnitz formula
$$(f(x)(x^3+x+2))^{(n)}=\\ f^{(n)}(x)(x^3+x+2)+nf^{(n-1)}(x)(3x^2+1)+\frac{n(n-1)}2f^{(n-2)}(x)(6x)+\frac{n(n-1)(n-2)}6f^{(n-3)}(x)(6),$$ you get the recurrence $$ 2f^{(n)}(0)+nf^{(n-1)}(0)+n(n-1)(n-2)f^{(n-3)}(0)=0.$$