$\begin{array}{l}\text{ Let }{e}^x=t,t\in (0,\infty )\\ \text{ Given equation will be }{t}^4+t^3-4t^2+t+1=0\\ \text{ Divide by }{t}^2\text{ to get }{t}^2+t-4+\frac{1}{t}+\frac{1}{t^2}=0\Rightarrow \left(t^2+\frac{1}{t^2}\right)+\left(t+\frac{1}{t}\right)-4=0\end{array}$

$\begin{array}{l}\text{ Let }t+\frac{1}{t}=\alpha ,\alpha \in [2,\infty )\Rightarrow t^2+\frac{1}{t^2}={\alpha }^2-2\\ \text{ So, the equation will be }\left({\alpha }^2-2\right)+\alpha -4=0\\ {\alpha }^2+\alpha -6=0\Rightarrow \alpha =-3,2\end{array}$

$\begin{array}{l}\because \alpha \in [2,\infty )\text{ ,we get }\alpha =2\Rightarrow e^x+e^{-x}=2\\ \text{ So, }x=0\text{ only solution }\end{array}$