Is there an analytical way to solve this? To be clear, I mean pow[x,x,x,x] = 2020.
Here is what I tried:
$$x^{x^{x^{x}}} = 2020$$
$x^{x^{x}} \ln x = \ln 2020$
${x^{x}}\ln x + \ln \ln x = \ln \ln 2020$
${(e^{\ln x})^{x}}\ln x + \ln \ln x = \ln \ln 2020$
${(e^{\ln x})^{e^{\ln x}}}\ln x + \ln \ln x = \ln \ln 2020$
Make $y = \ln x$
${(e^{y})^{e^{y}}}y + \ln y = \ln \ln 2020$
But I don´t know how to solve this equation.
By numerical methods I have $x \approx 1.94662238798$.
I do not think that a closed form could exist and, for sure, as uou did, a numerical method will be required.
Instead of looking for the zero of function $$f(x)=x^{x^{x^x}}-2020$$ which is varying so fast, search for the zero of $$g(x)=\log \left(\log \left(\log \left(x^{x^{x^x}}\right)\right)\right)-\log \left(\log \left(\log \left(2020\right)\right)\right)$$ which is quite smooth. A plot of $g(x)$ reveals that the solution is close to $x=2$.
Use one iteration of Newton method with $x_0=2$ and you will need to solve for $x$ $$\frac{(x-2) \left(8+64 \log ^3(2)+64 \log ^2(2)+32 \log (2)\right)}{\log (65536) \log (\log (65536))}+\log (\log (\log (65536)))=\log \left(\log \left(\log \left(2020\right)\right)\right)$$ Then, you have a nasty expression for $x_1$ which is $1.9448$. But continuing iterations, the pathe to solution will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.000000000000000000000000 \\ 1 & \color{red}{1.94}4787317029677396906107 \\ 2 & \color{red}{1.9466}19903004071435441212 \\ 3 & \color{red}{1.9466223879}75658343003294 \\ 4 & \color{red}{1.9466223879801945531843}44 \\ 5 & \color{red}{1.946622387980194553184359} \end{array} \right)$$
Let us see what higher order methods (Halley and Householder) would give $$\left( \begin{array}{cccc} n & \text{Newton} & \text{Halley} & \text{Householder} \\ 0 & 2.0000000000000000000 & 2.0000000000000000000 & 2.0000000000000000000 \\ 1 & 1.9447873170296773969 & 1.9465058701098434768 & 1.9466160957701974640 \\ 2 & 1.9466199030040714354 & 1.9466223879816314805 & 1.9466223879801945532 \\ 3 & 1.9466223879756583430 & 1.9466223879801945532 & \\ 4 & 1.9466223879801945532 & & \end{array} \right)$$
Inverse symbolic calculators do not identify the solution but the funny is that it is quite close to the reciprocal of the real root of the cubic $$39149 x^3-72556 x^2+71412 x-22845=0$$ which would give $20$ exact significant figures.