$\begin{array}{l} \left\{ \begin{array}{l} 2{x^2} + {y^2} - 2y - 2 = 0\\ {x^4} - 2{x^2}y = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {y^2} = 2y + 2 - 2{x^2}\left( 1 \right)\\ {x^4} - 2{x^2}{y^2} = 1\left( 2 \right) \end{array} \right.\\ \left( 1 \right) + \left( 2 \right) \Rightarrow {x^4} - 2{x^2}{y^2} + {y^2} = 2y - 2{x^2} + 3\\ \Leftrightarrow {\left( {{x^2} - y} \right)^2} = - 2\left( {{x^2} - y} \right) + 3\\ \Leftrightarrow {\left( {{x^2} - y} \right)^2} + 2\left( {{x^2} - y} \right) - 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l} {x^2} - y = 1\\ {x^2} - y = - 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x^2} = y + 1\\ {x^2} = y - 3 \end{array} \right.\\ + {x^2} = y + 1 \to \left( 1 \right)\\ \Rightarrow {y^2} = 2y + 2 - 2\left( {y + 1} \right) \Rightarrow {y^2} = 0 \Rightarrow y = 0\\ \Rightarrow {x^2} = 1 \Rightarrow x = \pm 1\\ + {x^2} = y - 3 \to \left( 1 \right)\\ \Rightarrow {y^2} = 2y + 2 - 2\left( {y - 3} \right) = 8 \Rightarrow y = \pm 2\sqrt 2 \\ \Rightarrow {x^2} = \pm 2\sqrt 2 - 3(Loại\, do x^2\ge 0)\\ \Rightarrow \left( {x;y} \right) = \left( {1;0} \right),\left( {-1; 0} \right) \end{array}$
