The equation to solve:
$$a^x+bf(x)+c = 0$$
Where $f(x)$ is a polynomial equation of degree $n$ without a constant term as it is covered by $c$.
I have solved the case where $f(x) = x$:
$$a^x+bx+c = 0$$
$$x_{1} = -\frac{c}{b} - \frac{W_{0}\big(\frac{(a^{\frac{-c}{b}})}{b}\big)}{ln(a)}$$
$$x_{2} = -\frac{c}{b} - \frac{W_{-1}\big(\frac{(a^{\frac{-c}{b}})}{b}\big)}{ln(a)}$$
$$a^x+bf(x)+c=0\tag{1}$$
$$a^x+bx^n+c=0$$
You already applied Lambert W for $n=1$.
a)
We can apply Lambert W also for $(n\neq 1) \land (c=0)$.
for real $a,b,x$; $k\in\{-1,0\}$:
$$a^x+bx^n=0$$ $$e^{\ln(a)x}+bx^n=0$$ $$e^{\ln(a)x}=-bx^n$$ $$1=-bx^ne^{-\ln(a)x}$$ $$-bx^ne^{-\ln(a)x}=1$$ $$x^ne^{-\ln(a)x}=-\frac{1}{b}$$ $$xe^{-\frac{\ln(a)}{n}x}=\left(-\frac{1}{b}\right)^\frac{1}{n}$$ $$-\frac{\ln(a)}{n}xe^{-\frac{\ln(a)}{n}x}=-\frac{\ln(a)}{n}\left(-\frac{1}{b}\right)^\frac{1}{n}$$ $$-\frac{\ln(a)}{n}x=W_k\left(-\frac{\ln(a)}{n}\left(-\frac{1}{b}\right)^\frac{1}{n}\right)$$ $$x=-\frac{n}{\ln(a)}W_k\left(-\frac{\ln(a)}{n}\left(-\frac{1}{b}\right)^\frac{1}{n}\right)$$
b)
Your general equation (1) can be solved in terms of Generalized Lambert W.