For approximating the Lambert W function, the Wikipedia page lists Newton's method and Halley's method.
Newton's method (the Newton–Raphson method) converges quadratically (number of correct digits doubles per iteration). For approximating the Lambert W function, it is defined as:
$$w_{j+1}=w_j-\frac{w_je^{w_j}-z}{e^{w_j}+w_je^{w_j}}$$
Halley's method converges cubically (number of correct digits triples per iteration). For approximating the Lambert W function, it is defined as:
$$w_{j+1}=-\frac{w_je^{w_j}-z}{e^{w_j}(w_j+1)-\frac{(w_j+2)(w_je^{w_j}-z)}{2w_j+2}}$$
Do other methods exist that converge faster than these? If so, what are they?
I have thus far been unable to find any that are used to approximate the Lambert W function.
If you look for higher order formulae, you could consider Householder method which is or order $4$.
If you want to go further, have a look at this paper; using it, you can easily build an iteration which would be of the order you wish.
The equation to solve is $$f(w)=w e^w-z=0\implies f^{(n)}=e^w(w+n)$$ which makes the expression quite simple for the iterates.
For illustration purposes, let us consider $z=123456789$ and, being lazy, use $w_0=\log(123456789)\approx 18.6314$. The following table reports the first iterate $w_1$ as a function of the order $k$ of the method $$\left( \begin{array}{ccc} k & w_1 & \text{method} \\ 2 & 17.73327936 & \text{Newton} \\ 3 & 17.30942321 & \text{Halley}\\ 4 & 17.04239937 & \text{Householder}\\ 5 & 16.85306931 & \text{no name}\\ 6 & 16.70984559 & \text{no name}\\ 7 & 16.59697020 & \text{no name}\\ 8 & 16.50546219 & \text{no name}\\ 9 & 16.42972594 & \text{no name} \end{array} \right)$$ while the exact solution is $15.86715078$; it is quite slow because the starting value $w_0$ is not good enough.
Using a better approximation, as shown here, that is to say $$W(z)\approx L_1-L_2+\frac{L_2}{L_1} \qquad \qquad L_1=\log(z)\qquad L_2=\log(L_1)$$ for the chosen example $(w_0\approx 15.8635)$, the same methods will give $$\left( \begin{array}{ccc} k & w_1 & \text{method} \\ 2 & 15.867157703589425302 & \text{Newton}\\ 3 & 15.867150764845732185 & \text{Halley}\\ 4 & 15.867150782609469895 & \text{Householder}\\ 5 & 15.867150782558279240 & \text{no name}\\ 6 & 15.867150782558436641 & \text{no name}\\ 7 & 15.867150782558436137 & \text{no name}\\ 8 & 15.867150782558436139 & \text{no name}\\ 9 & 15.867150782558436139 & \text{no name} \end{array} \right)$$ while the exact solution is $15.867150782558436139$.
Any way, my personal opinion is that, for $z>0$, it is simpler to find the zero of $$g(w)=\log(w)+w-\log(z)$$ and apply any method to it since the derivatives $g^{(n)}$ are very simple and the $a_i$ are just very simple.
To compare the behavior of the two functions, I show below Newton iterations for the previous value of $z$ still starting with the poorest estimate $$\left( \begin{array}{ccc} n & \text{using f(w)} & \text{using g(w)} \\ 1 & 17.7332793612487 & 15.8555416036877 \\ 2 & 16.9177099266940 & 15.8671505305371 \\ 3 & 16.2832371769982 & 15.8671507825584 \\ 4 & 15.9466743519793 & \\ 5 & 15.8704075861194 & \\ 6 & 15.8671563938690 & \\ 7 & 15.8671507825751 & \\ 8 & 15.8671507825584 & \end{array} \right)$$