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

Đặt ${{4}^{x+y}}={{3}^{{{x}^{2}}+{{y}^{2}}}}=t$, $t>0$ $\Leftrightarrow \left\{ \begin{aligned}
& x+y={{\log }_{4}}t \\
& {{x}^{2}}+{{y}^{2}}={{\log }_{3}}t \\
\end{aligned} \right.$.
Vì ${{\left( x+y \right)}^{2}}\le 2\left( {{x}^{2}}+{{y}^{2}} \right)\Rightarrow \log _{4}^{2}t\le 2{{\log }_{3}}t\Leftrightarrow \dfrac{{{\ln }^{2}}t}{{{\ln }^{2}}4}\le 2\dfrac{\ln t}{\ln 3}\Leftrightarrow 0\le \ln t\le \dfrac{2{{\ln }^{2}}4}{\ln 3}$.
Suy ra ${{x}^{2}}+{{y}^{2}}=\dfrac{\ln t}{\ln 3}\le \dfrac{2{{\ln }^{2}}4}{{{\ln }^{2}}3}=2{{\left( \dfrac{\ln 4}{\ln 3} \right)}^{2}}\approx 3,18\Rightarrow {{x}^{2}}\le 3,18\xrightarrow{x\in \mathbb{Z}}x\in \left\{ -1;0;1 \right\}$.
Nếu $x=0\Rightarrow \left\{ \begin{aligned}
& 0+y={{\log }_{4}}t \\
& {{0}^{2}}+{{y}^{2}}={{\log }_{3}}t \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& y=0 \\
& t=1 \\
\end{aligned} \right.$.
Nếu $x=1\Rightarrow \left\{ \begin{aligned}
& 1+y={{\log }_{4}}t \\
& {{1}^{2}}+{{y}^{2}}={{\log }_{3}}t \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& y=\dfrac{\ln t}{\ln 4}-1 \\
& {{\left( \dfrac{\ln t}{\ln 4}-1 \right)}^{2}}+1=\dfrac{\ln t}{\ln 3} \\
\end{aligned} \right.$$\Rightarrow \exists t\Rightarrow \exists y$.
Nếu $x=-1\Rightarrow \left\{ \begin{aligned}
& -1+y={{\log }_{4}}t \\
& {{\left( -1 \right)}^{2}}+{{y}^{2}}={{\log }_{3}}t \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& y=\dfrac{\ln t}{\ln 4}+1 \\
& {{\left( \dfrac{\ln t}{\ln 4}+1 \right)}^{2}}+1=\dfrac{\ln t}{\ln 3} \\
\end{aligned} \right.$$\Rightarrow \not{\exists }t\Rightarrow \not{\exists }y$.
Vậy $x\in \left\{ 0;1 \right\}$.