$(2 \pm \sqrt{\mathfrak{problem}})^x$

Notice that $\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$ and $7-4\sqrt{3} =(2-\sqrt{3})^2$ .Then we have $\begin{aligned} (2- \sqrt{3})^x+(7-4 \sqrt{3})(2+ \sqrt{3})^x & =4(2- \sqrt{3}) \\ \implies (2-\sqrt{3})^x+\frac{7-4\sqrt{3}}{(2-\sqrt{3})^x} & =4(2-\sqrt{3}) \\ \implies (2-\sqrt{3})^{2x}-4(2-\sqrt{3})(2-\sqrt{3})^x+7-4\sqrt{3} & = 0 \quad \quad [\text{Multiplying both sides by }~(2-\sqrt{3})^x] \\ \implies a^2-4(2-\sqrt{3})a+7-4\sqrt{3} & =0 \quad \quad [\text{where}~ a=(2-\sqrt{3})^x] \end{aligned}$ The above equation being a quadratic equation,has it roots given by the quadratic formula as $\begin{aligned} a &= \frac{4(2-\sqrt{3})\pm\sqrt{\{4(2-\sqrt{3})\}^2-4(7-4\sqrt{3}})}{2} \\ & =\frac{8-4\sqrt{3}\pm(2\sqrt{3}\sqrt{7-4\sqrt{3}})}{2} \\ & =4-2\sqrt{3}\pm(2\sqrt{3}-3)\quad \quad [\because (7-4\sqrt{3})=(2-\sqrt{3})^2] \\ \implies a & =1,(7-4\sqrt{3}) \implies (2-\sqrt{3}) =1,(2-\sqrt{3})^2 \\ \therefore x & =\boxed{0,2} \end{aligned}$

Hence,the only non-zero root is $\displaystyle \boxed{\color{#20A900}{2}}$ .

Just re-posting the solution given by @Sabhrant Sachan

$(2-\sqrt3)^x+(2-\sqrt3)^{2-x}=4(2-\sqrt3) \\ (2-\sqrt3)^{x-1}+(2-\sqrt3)^{1-x}=4 \\ a = (2-\sqrt3)^{x-1} \\ a+\dfrac1{a}=4 \\ a^2-4a+1=0 \\ a = \dfrac{4\pm\sqrt{16-4}}{2} \\ a=2\pm\sqrt3 \\ (2-\sqrt3)^{x-1} = 2-\sqrt3 , \dfrac{1}{2-\sqrt3} \\ x-1=1 \text{ or }-1 \\ \implies x=2 \text{ or } 0$