Có bao nhiêu số tự nhiên $x$ sao cho mỗi giá trị x tồn tại số $y$...

Điều kiện:
Đặt , suy ra $\left\{ \begin{aligned}
& x-y={{3}^{t}} \\
& {{x}^{2}}+2{{y}^{2}}\le {{6}^{t}} \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x={{3}^{t}}+y \\
& {{\left( {{3}^{t}}+y \right)}^{2}}+2{{y}^{2}}\le {{6}^{t}}\begin{matrix}
{} & \left( 1 \right) \\
\end{matrix} \\
\end{aligned} \right.\left( 1 \right)\Leftrightarrow 3{{y}^{2}}+{{2.3}^{t}}y+{{9}^{t}}-{{6}^{t}}\le 0{\Delta }'={{9}^{t}}-3\left( {{9}^{t}}-{{6}^{t}} \right)\ge 0\Leftrightarrow {{\left( \dfrac{2}{3} \right)}^{t}}\ge \dfrac{2}{3}\Leftrightarrow t\le 1{{x}^{2}}+2{{y}^{2}}\le 6\Rightarrow {{x}^{2}}\le 6\Rightarrow x\in \left\{ 0;1;2 \right\}x\in \mathbb{N}x=0\Rightarrow \left\{ \begin{aligned}
& y=-{{3}^{t}} \\
& 2{{y}^{2}}\le {{6}^{t}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t\le {{\log }_{\dfrac{2}{3}}}2<0 \\
& y=-{{3}^{t}}\in \left( -1;0 \right) \\
\end{aligned} \right.x=1\Rightarrow \left\{ \begin{aligned}
& 1-y={{3}^{t}} \\
& 1+2{{y}^{2}}\le {{6}^{t}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& y=1-{{3}^{t}} \\
& 1+2{{\left( 1-{{3}^{t}} \right)}^{2}}\le {{6}^{t}} \\
\end{aligned} \right. t=0,y=0x=2\Rightarrow \left\{ \begin{aligned}
& 2-y={{3}^{t}} \\
& 4+2{{y}^{2}}\le {{6}^{t}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& y=2-{{3}^{t}} \\
& 4+2{{\left( 2-{{3}^{t}} \right)}^{2}}\le {{6}^{t}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& y=2-{{3}^{t}} \\
& {{9}^{t}}-{{6}^{t}}-{{8.3}^{t}}+12\le 0 \\
\end{aligned} \right. t=1,y=-1x\in \left\{ 0;1;2 \right\}$.