$\begin{array}{l} \left\{ \begin{array}{l} {x^2} + xy + {y^2} - 4y + 1 = 0\\ \left( {{x^2} + 1} \right)\left( {x + y - 2} \right) = y \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} + xy + {y^2} + 1 = 4y\left( 1 \right)\\ \left( {{x^2} + 1} \right)\left( {x + y - 2} \right) = y\left( 2 \right) \end{array} \right.\\ \left( 2 \right) - \left( 1 \right) \Rightarrow \left( {{x^2} + 1} \right)\left( {x + y - 2} \right) - \left[ {\left( {{x^2} + 1} \right) + xy + {y^2}} \right] = - 3y\\ \Leftrightarrow \left( {{x^2} + 1} \right)\left( {x + y - 3} \right) = {y^2} + xy - 3y\\ \Leftrightarrow \left( {{x^2} + 1} \right)\left( {x + y - 3} \right) = y\left( {x + y - 3} \right)\\ \Leftrightarrow \left( {x + y - 3} \right)\left( {{x^2} + 1 - y} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + y - 3 = 0\\ {x^2} + 1 = y \end{array} \right.\\ + {x^2} + 1 = y \Rightarrow \left( 2 \right):y\left( {x + y - 3} \right) = y\\ \Rightarrow x + y - 3 = 0\left( {do\,y = {x^2} + 1 \ge 1 > 0} \right)\\ \Rightarrow y = 3 - x\\ \Leftrightarrow {x^2} + 1 = 3 - x \Leftrightarrow {x^2} + x - 2 = 0 \Leftrightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = - 2 \Rightarrow y = 5\\ x = 1 \Rightarrow y = 3 \end{array} \right. \end{array}$
$x+y-3=0$. Lặp lại trường hợp trường hợp trên ta cũng thu được nghiệm tương tự.
Vậy $\left( {x;y} \right) = \left( {1;2} \right),\left( { - 2;5} \right)$
