How to check if some equation can be solved using Lambert $\operatorname{W}$ function.
Assume an ordinary equation $F(x)=c$ is given where $c$ is a constant and $F$ is a function. Isolating $x$ on one side of the equation only by operations to the whole equation means to apply a suitable partial inverse function (branch of the inverse relation) $F^{-1}$ of $F$: $\ x=F^{-1}(c)$.
The problem of existence of elementary inverses of elementary functions is solved by the theorem in [Ritt 1925] that is proved also in [Risch 1979].
The problem of existence of elementary numbers as solutions of irreducible polynomial equations $P(x,e^x)=0$ is solved in [Lin 1983] and [Chow 1999].
But LambertW is not an elementary function.
I. a. the following kinds of equations of $x$ can be solved in closed form by applying Lambert W or without Lambert W.
Let
$c_1,...,c_8\in\mathbb{C}$,
$f,f_1$ functions in $\mathbb{C}$ with suitable local closed-form inverses.
$$\tag 1 c_1x^{c_2}+c_3x^{c_4}\left(e^{c_5+c_6x^{c_7}}\right)^{c_8}=0$$
$$\tag 2 c_1f(x)^{c_2}+c_3f(x)^{c_4}\left(e^{c_5+c_6f(x)^{c_7}}\right)^{c_8}=0$$
If $c_2,c_4\neq0$, $x=0$ and $f(x)=0$ respectively is a solution.
$$\tag 3 c_1+c_2x+e^{c_3+c_4x}=0$$
$$\tag 4 f_1\left(c_1+c_2f(x)+e^{c_3+c_4f(x)}\right)=0$$
If you have an $x$-containing summand on the left-hand side of the equation, you can subtract the exponential term from the equation and divide the equation by it to get a product of a power of $x$ (or $f(x)$) and an exponential term of $x$ (or $f(x)$) to apply Lambert W.
See also [Edwards 2020], [Galidakis/ Weisstein], [Köhler].
If your equation contains an exponential function or a logarithm function but cannot be brought to the form of equations (1) - (4), you could try to apply a predefined generalization of Lambert W.
See e.g. [Corcino/Corcino/Mezö 2017], [Dubinov/Galidakis 2007], [Galidakis 2005], [Maignan/Scott 2016], [Mezö 2017], [Barsan 2018].
$\ $
[Barsan 2018] Barsan, V.: Siewert solutions of transcendental equations, generalized Lambert functions and physical applications. Open Phys. 16 (2018) 232–242
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
[Corcino/Corcino/Mezö 2017] Corcino, C. B.; Corcino, R. B.; Mezö, I.: Integrals and derivatives connected to the r-Lambert function. Integral Transforms and Special Functions 28 (2017) (11)
[Dubinov/Galidakis 2007] Dubinov, A.; Galidakis, Y.: Explicit solution of the Kepler equation. Physics of Particles and Nuclei Letters 4 (2007) 213-216
[Edwards 2020] Edwards, S.: Extension of Algebraic Solutions Using the Lambert W Function. 2020
[Galidakis 2005] Galidakis , I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997
[Galidakis/Weisstein] Galidakis, I.; Weisstein, E. W.: Power Tower. Wolfram MathWorld
[Köhler] Köhler, Th: Gebrauch der Lambertschen W-Funktion (Omegafunktion)
[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
[Maignan/Scott 2016] Maignan, A.; Scott, T. C.: Polynomial-Exponential and Generalized Lambert Function. 2016
[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
Generally, you can't solve something in the following form:$$x^{x+c}$$because of the stupid "$c$".
If you can't get it into the form $f(x)e^{f(x)}$, it is also probably not solvable.
You usually can't solve $$f(x)g(x)$$where $f(x)$ is more than one "level" away from $g(x)$. For example:$$xe^{e^x}$$has no solution because $x$ is two "levels" away from $e^{e^x}$.
Negative levels can be treated as logarithms.
The Lambert W function does not like addition, which is why you usually cancel addition by exponentiation and applying exponent properties.
Any base is considered fine (usually) because you can change the base with exponent/logarithm properties as well.