Approximate solution of $H(x)=(x!)^k$

My post being already too long, I prefer to add an answer to it rather than to edit it.

I started with a different approach considering instead that I am looking for the zero of function $$g(x)=\log (H(x))-k \log (x!)$$ Using $x_0=2k$, the first iterate of Newton method is given by $$x_1=2k-\frac{2 \log (H(2 k))-2 k \log ((2k)!)}{2 \log ((2k)!)+4 k-2 k \psi(2 k+1)+1-\log (2 \pi )}$$ Now, using expansion for large values of $k$, I end with $$\color{blue}{x_1^*= 2 k-\frac{2 k+\log (2 k)-\log (2 \pi )}{2 \log (2 k)}+\frac{2 \log (2 k)+1}{4 \log ^2(2 k)}}$$

For the test example $(k=1234)$, this gives $x_1=2309.706772$, $x_1^*=2309.706724$ while the exact solution is $2297.186319$.

From a numerical point of view, it is better to consider $g(x)$ rather than $f(x)$ since, for any of the proposed guesses $x_*$ $f(x_*) \, f''(x_*) <0$ means that, by Darboux theorem, we should have one overshoot of the solution while $g(x_*) \, g''(x_*) <0$ guarantees no overshoot at all.

Update

Doing the same work with the approximating function mentioned in the post

$$h(x)=(2 \log (x)-1)x-{4k( \log (x)-1)}$$ which is much simpler to handle, using high order methods, I obtained things such $$x_2=2k \frac{\sum_{i=1}^9 a_i t^i}{\sum_{i=1}^9 b_i t^i} \qquad \text{where} \qquad t={\log(2k)}$$

The $a_i$'s correspond to the sequence $$\{667,-111648,-156240,1348032,803040,-4435200,564480,4515840,-3225600,645120\}$$

and the $b_i$'s to $$\{-4509,-78720,69552,1099392,-272160,-3548160,1693440,3225600,-2903040,645120\} $$

From those, truncated series could easily be obtained by long division; for example $$x_2=2k\left(1-\frac{1}{2 t}-\frac{1}{4 t^2}-\frac{3}{8 t^3}-\frac{1}{2 t^4}-\frac{77}{96t^5}+O\left(\frac{1}{t^6}\right)\right)$$

Using the terms of the present table, for the test case we should get $x_2=2297.54854638212$ while the solution of $h(x)=0$ is $2297.54854638189$ and the exact solution of $g(x)=0$ is $2297.186319$.