It is easy to notice that

$9^{x+1}+2187=3^{6x-x^2}$

$ \implies 3^{2x+2}+3^7=3^{6x-x^2}$

$\implies 3^{x^2-4x+2}+3^{x^2-6x+7}=1$

$\implies 3^{x^2-4x+4}+3^{x^2-6x+9}=9$

$\implies 3^{(x-2)^2}+3^{(x-3)^2}=9$

so it suffices to prove the following claim:

If $x=3-\sqrt{2}+k$ is a solution, then $x=2+\sqrt{2}-k$ is a solution, for an arbitrary real number $k$.