We claim the answer is $\boxed{5}$. 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 $k$. $\textit{Proof:}$ This is true by simply substituting into the last equation and checking: $x= 3-\sqrt{2}+k \implies 3^{(k-\sqrt{2})^2}+3^{(k+1-\sqrt{2})^2}=9$ $x= 2+\sqrt{2}-k \implies 3^{(\sqrt{2}-k)^2}+3^{(\sqrt{2}-k-1)^2}=9$ The RHS of the two above equations are clearly equivalent, so the claim is true. $\square$ Thus, it suffices to choose minimal or maximal $k$, and the sum will hence be $5$, as desired. $\blacksquare$