Show that $\frac{1+z-\sqrt{z^2-6z+1}}{4}$ fits the Lagrangean framework

Let $S(z)$ be the OGF of bracketings. Show that the Lagrangean framework holds for $S(z)$.

Remark: You can find the definition of Lagrangean framework below.

From the Flajolet & Sedgewick book (p.81) we know that for $S(z)$ holds

$$S(z) = \frac{1+z-\sqrt{z^2-6z+1}}{4} \quad \text{ and } \quad S(z) = z + \frac{S(z)^2}{1-S(z)}.$$

and according to Wolfram Alpha the series expansion of $S(z)$ looks like this:

$$S(z) = z + z^2 + 3 z^3 + 11 z^4 + 45 z^5 + 197 z^6 + 903 z^7 + 4279 z^8 + 20793 z^9 + 103049 z^{10} + 518859 z^{11} + \mathcal{O}(z^{12}).$$

However, I do not know how to rigorously find a formula for the coefficients of $\phi(z)$ to show condition $(A)$. Could you please give me a hint?

Lagrangean Framework: We say that $f$ fulfills the Lagrangean framework if$$f(z) = z \phi(f(z)), \qquad (A)$$where both $\phi$ and $f$ are formal power series and if the following conditions for $\phi(z) := \sum_{j \ge 0} a_z z^j$ hold:

1. $\phi(0) \ne 0$
2. (Nonnegativity)$a_j \ge 0 \quad (j \ge 0)$,
3. (Aperiodicity)$gcd \{j : a_j > 0 \} = 1$;
4. (Analyticity)$\phi$ is analytic in $\lvert z \rvert < R, 0 < R \le \infty$;
5. (Sub-criticality)$r\phi^\prime(r) = \phi(r)$ for some $r \in (0,R)$

EDIT: Folllowing anon's suggestion I computed

$$\phi(z) = 1 + \sum_{n \ge 1} 2^{n-1} z^n$$

Below you can find some intermediate results:

\begin{align}S(z) &= z + z^2 + 3 z^3 + 11 z^4 + 45 z^5 + 197 z^6 + 903 z^7 + 4279 z^8 + 20793 z^9 + 103049 z^{10} + 518859 z^{11} + \mathcal{O}(z^{12}) \\S(z)^2 &= z^2 + 2 z^3 + 7 z^4 + 28 z^5 + 121 z^6 + 550 z^7 + \mathcal{O}(z^8)\\S(z)^3 &= z^3 + 3 z^4 + 12 z^5 + 52 z^6 + 237 z^7 + 1119 z^8 + \mathcal{O}(z^9)\\S(z)^4 &= z^4 + 4 z^5 + 18 z^6 + 84 z^7 + 403 z^8 + 1976 z^9 + 9860 z^{10} + \mathcal{O}(z^{11}) \\S(z)^5 &= z^5 + 5 z^6 + 25 z^7 + 125 z^8 + 630 z^9 + 3206 z^{10} + \mathcal{O}(z^{11})\end{align}

However, I do not see how to formally prove the identity

$$S(z) = z \phi(S(z))$$

Could you please give me a hint?