Solutions are (2,3) and (-3,-2) and reverses. You can let x^2+y2^=u, xy=v and substitute. Then wlog let x=y+1 due to cyclicity and solve u get v=6

$$| x - y | = 1$$$$\implies x^2 + y^2 = 1 + 2xy$$$$(x-y)^2 = (x+y)^2 - 4xy$$$$\implies (x+y)^2 = 1 + 4xy$$$$\text{Let} \quad xy = a$$$$(x^{2}+y^{2})^{2}-xy(x+y)^{2}=19$$$$\implies (1 + 2xy)^2 - xy(1 + 4xy) = 19$$$$\implies (2a + 1)^2 - a(4a + 1) = 19$$$$\implies a = 6$$Wlog $x > y$ So we have $xy = 6$ and $x-y =1$ So $(x,y) = (3,2),(-2,-3)$ and since we assumed Wlog hence the other 2 solutions are $(2,3) , (-3,-2)$

Find all the pairs of real numbers $(x,y)$ that are solutions of the system: $$(x^{2}+y^{2})^{2}-xy(x+y)^{2}=19 $$$$| x - y | = 1$$ $$\textbf{Case 1:} x-y=1$$ Then $x=1+y$, the equation becomes $$ \left( (1+y)^2+y^2 \right)^2-(1+y)(y)(1+y+y)^2=19 \leftrightarrow$$$$(2y^2+2y+1)^2-(y^2+y)(2y+1)^2=19 \leftrightarrow$$$$3y^2 + 3y + 1=19 \leftrightarrow $$$$3y^2+3y-18=0$$$$y=\frac{-3 \pm \sqrt{9+12\cdot18}}{6}=\frac{-3 \pm 15}{6}$$$$y=2,-3$$$$\textbf{Case 2:} y-x=1$$Then $y=1+x$ , Observe that it is the same thing but different order as the Case 1., cuz swapping x and y in the original equation does not change anything at all, hence only solutions are $(y,x) \in (2,3),(-3,-2)$ and its permutations

WLOG x-y=1, solve for y and x, which are 3, 2 and -2, -3, because all of the degree of x's greater than 2 cancel out, so it is very convenient that we can just solve for the quadratic. Also, if y-x=1, the eq. is symmetric so we just swap them giving 2, 3 and -3, -2