Find number of solutions of the equation $x^4-1=e^x$.

I tried making a rough plot and trying to find number of intersections and got that either number of solutions will be $1$ or $3$. But I am confused how to check which is the case out of the following two:

This is graph of $x^2-1$ and $e^x$ which give only one solution.

And below is actual graph of $x^4-1$ which $e^x$:

So I want to know how to decide which graph will be the case here without putting values of $x$ and check(as it works here as small values but may not in other cases). I want to know a general approach. Like if we have to find solutions of $x^{12}-1 = e^{3.25x}$, how do we know what will be the case.

Here's a hand-wavy idea that someone can flesh out if they please.

Given a positive integer $n$ and positive real number $a$, the task is to find out how many solutions $x^{2n} -1 = e^{ax}$ has. Replace $e^{ax}$ by its Maclauren polynomial of sufficient degree subtract the left side from the right:

$$2 + ax +\frac{(ax)^2}{2!} + \cdots + \left(\frac{a^{2n}}{(2n)!}-1\right){x^{2n}} +\cdots = 0.$$

By Descartes'Rule of Signs, if $\frac{a^{2n}}{(2n)!}-1$ is positive, there are no sign changes and hence no positive zeroes. If it's negative then there are two sign changes and so there might be two zeroes.

I checked this out on the suggested $x^{12} - 1=e^{3.25x}$. There are 3 solutions in this case and we have $\frac{a^{2n}}{(2n)!}-1 = -.997\ldots$. Then I tried $x^{12} -1= e^{6x}.$ In this case $\frac{a^{2n}}{(2n)!}-1 = 3.544\ldots$, and the graph shows there is only one solution.

So I believe that the sign of $\frac{a^{2n}}{(2n)!}-1$ is a good indicator of whether there are one or three zeroes.