Pirkuliyev Rovsen wrote: Solve in natural the equation $9^x-3^x=y^4+2y^3+y^2+2y$ $3^x(3^x-1)=y(y+2)(y^2+1)$ $y^2+1\not\equiv 0\pmod 3$ and $3$ cant divide both $y$ and $y+2$ and so $3^x|y$ or $3^x|y+2$. But $3^x|y$ implies $y\ge 3^x$ and so $(y+2)(y^2+1)\le 3^x-1<y$ impossible And $3^x|y+2$ implies $y+2\ge 3^x$ and so $y(y^2+1)\le 3^x-1<y+2$ and so $y=1$ and then $x=1$ which indeed is a solution Hence the unique natural solution : $\boxed{(x,y)=(1,1)}$

Another idea: Find positive integer $a$, in terms of $y$, so that the inequality $a^2-a<y^4+2y^3+y^2+2y<(a+1)^2-(a+1)$ is true for almost all $y$