Từ pt `(2)` ta có

`x^3 + 3(x^2+y^2) + 4x = y^3 + 4y-4`

`<=> x^3 + 3x^2 + 3y^2 + 4x= y^3 + 4y - 4`

`<=>x^3 + 3x^2 + 4x+4 = y^3+4y - 3y^2`

`<=> x^3 + 3x^2+3x+1 + x + 1 + 2 = y^3 - 3y^2+3y-1 + y- 1+2`

`<=> (x+1)^3 + (x+1) + 2 = (y-1)^3 + (y-1) + 2`

`<=> (x+1)^3 + (x+1) = (y-1)^3 + (y-1)`

`<=> [(x+1)^3 - (y-1)^3] + (x+1-y+1) = 0`

`<=>(x-y+2)[(x+1)^2 + (x+1)(y-1) + (y-1)^2] + (x-y+2) = 0`

`<=> (x-y+2)[(x+1)^2 + (x+1)(y-1) + (y-1)^2+1] = 0`

`th1`

`x-y+2 = 0`

`<=> y = x+2`

Thay vào pt `(1)` ta đc

`505x + 253(x+2) = 2022`

`<=> 505x + 253x + 506 = 2022`

`<=> 758x =1516`

`<=> x=2`

`=> y= 4`

`th2`

`(x+1)^2 + (x+1)(y-1) + (y-1)^2+1 = 0`

Đặt `x+1 = a; y-1 = b` 

Ta đc

`a^2 + ab + b^2 = 0`

`a^2 + 2.a.b/2 + b^2/4 + {3b^2}/4 + 1 = 0`

`(a+b/2)^2 + {3b^2}/4 + 1 = 0`

Mà `(a+b/2)^2 + {3b^2}/4 + 1 > 0`

`=> pt` vô nghiệm

Vậy hpt có nghiệm `(x;y) = (2;4)`