Question: Let $w=t(1+w)^z$, and $$G(t)=1+\sum_{k\geq1}\frac{x}{k}\binom{x-1+kz}{k-1}t^k.$$ Prove that $G(t)=(1+w)^x$.
My attempt: From the first implicit functional equation I can use the Lagrange Implicit Function Theorem (LIFT) to solve for $w$ as a formal power series in $t$. In particular each coefficient of this power series is given by, $$[t^k]w(t)=\frac{1}{k}[\lambda^{k-1}]\left((1+\lambda)^z\right)^k=\frac{1}{k}\binom{zk}{k-1}.$$ I could then use this formula for $w(t)$ and try to expand $(1+w(t))^x$ and show this is equal to $G(t)$. However, expanding gives $$\left(1+\sum_{k\geq 0}\frac{1}{k}\binom{zk}{k-1}t^k\right)^x=\sum_{m\geq 0}\binom{x}{m}\left(\sum_{k\geq 0}\frac{1}{k}\binom{zk}{k-1}t^k\right)^m,$$ which just seems to go nowhere...
Alternatively, if I could find a power series $F\left((1+w)^x\right)=(1+x)^z$ would it then be sufficient to check that $w(t)=tF((G(t))$, or is there something more clever I can do, perhaps I am missing some consequence of LIFT that would make checking this easy?
We start by observing that $G(0)=1.$ Next we use the notation from Wikipedia on Lagrange-Buermann and note that $\phi(w) = (1+w)^z$ and $H(w) = (1+w)^x$ to get for $k\ge 1$
$$[t^k] (1+w)^x = \frac{1}{k} [w^{k-1}] (H'(w) \phi(w)^k) = \frac{1}{k} [w^{k-1}] x (1+w)^{x-1} (1+w)^{kz} \\ = \frac{x}{k} [w^{k-1}] (1+w)^{x-1+kz} = \frac{x}{k} {x-1+kz\choose k-1}.$$
If we want to derive this from first principles we may use the Cauchy Coefficient Formula. We have $A(t) = t (1+A(t))^z$ and seek $[t^k] (1+A(t))^x.$ To start we observe that
$$[t^k] (1+A(t))^x = \frac{1}{k} [t^{k-1}] x (1+A(t))^{x-1} A'(t)$$
which is
$$\frac{x}{k} \frac{1}{2\pi i} \int_{|t|=\epsilon} \frac{1}{t^k} (1+A(t))^{x-1} A'(t) \; dt.$$
Putting $A(t) = w$ we find
$$\frac{x}{k} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{kz}}{w^k} (1+w)^{x-1} \; dw \\ = \frac{x}{k} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{x-1+kz}}{w^k} \; dw = \frac{x}{k} {x-1+kz\choose k-1}.$$
Reference for this is Analytic Combinatorics by Flajolet and Sedgewick, p. 732, Lagrange Inversion.