Given,

$e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0$

$\Rightarrow e^{4x}-e^{2x}-3-\frac{1}{e^{2x}}+\frac{1}{e^{4x}}=0$

$\Rightarrow e^{4x}+\frac{1}{e^{4x}}+2-\left(e^{2x}+\frac{1}{e^{2x}}\right)=5$

$\Rightarrow {\left(e^{2x}+\frac{1}{e^{2x}}\right)}^2-\left(e^{2x}+\frac{1}{e^{2x}}\right)=3$

Now let $\left(e^{2x}+\frac{1}{e^{2x}}\right)=t$

So, the equation becomes

$t^2-t-5=0$

$\Rightarrow t=\frac{1+\sqrt{21}}{2}$, ignoring negative sign as exponential function are positive,

Now $e^{2x}+\frac{1}{e^{2x}}=\frac{1+\sqrt{21}}{2}$

$\Rightarrow e^{4x}-\frac{1+\sqrt{21}}{2}{e}^{2x}+1=0$ which is quadratic equation in $e^{2x}$ with upward parabola, 

Now by $A.M\ge G.M$ we get, $e^{2x}+\frac{1}{e^{2x}}\ge 2$

So, $y=\frac{1+\sqrt{21}}{2}>2$, hence it will cut at two distinct point,

Hence, there will be two solution.