Có bao nhiêu số nguyên $x$ sao cho tồn tại số thực $y$ thoả mãn...

Đặt: $2{{\log }_{3}}\left( x+y+1 \right)={{\log }_{2}}\left( {{x}^{2}}+2x+2{{y}^{2}}+1 \right)=t$.
$\Leftrightarrow \left\{ \begin{aligned}
& x+y+1={{\left( \sqrt{3} \right)}^{t}} \\
& {{x}^{2}}+2x+2{{y}^{2}}+1={{2}^{t}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x+y+1={{\left( \sqrt{3} \right)}^{t}} \\
& {{\left( x+1 \right)}^{2}}+2{{y}^{2}}={{2}^{t}} \\
\end{aligned} \right.$.
Đặt $\left\{ \begin{aligned}
& X=x+1 \\
& Y=\sqrt{2}y \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& X+\dfrac{Y}{\sqrt{2}}={{\left( \sqrt{3} \right)}^{t}} \\
& {{X}^{2}}+{{Y}^{2}}={{2}^{t}} \\
\end{aligned} \right. $có nghiệm $ \Rightarrow d\left( O,\Delta \right)\le R$
$\Rightarrow \dfrac{\left| -{{\left( \sqrt{3} \right)}^{t}} \right|}{\sqrt{{{1}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}}\le {{\left( \sqrt{2} \right)}^{t}}\Leftrightarrow {{\left( \sqrt{3} \right)}^{t}}\le \dfrac{\sqrt{6}}{2}.{{\left( \sqrt{2} \right)}^{t}}\Leftrightarrow {{\left( \dfrac{\sqrt{6}}{2} \right)}^{t}}\le \dfrac{\sqrt{6}}{2}\Rightarrow t\le 1$
$\begin{aligned}
& \Rightarrow {{X}^{2}}+{{Y}^{2}}\le 2 \\
& \Rightarrow \left| X \right|\le 2\Leftrightarrow -\sqrt{2}\le x+1\le \sqrt{2}\Leftrightarrow -2,41\le x\le 0 \\
\end{aligned}$
Mà $x$ nguyên $\Rightarrow x=-2;-1;0$.