$P(x)=e^x$ has a finite number of solutions . [duplicate]
An alternative solution would be to first note that, the set of solutions must be bounded.
This is because for some $M_1$, you have $e^x>P(x)$ for all $x>0$, and some $M_2$ such that $P(x)>1$ for $x<M_2$ (because the limit of $P$, as $x\to-\infty$, is $\pm \infty$.
Because the set of solutions is bounded, it must have a convergence point. This is not possible because the function $g(x)=P(x)-e^x$ is analytic and non-constant, and the roots of such function are always isolated.