Someone correct me if I did something wrong. $$y^5+2xy=x^2+2y^4$$$$\Leftrightarrow y^5-2y^4+y^2 = x^2 -2xy +y^2$$$$\Leftrightarrow y^2(y^3-2y^2+1) = (x-y)^2$$ Hence $y^3-2y^2+1 = a^2$ for some $a \in \mathbb{Z}$ implying $y \geq 0$. $(y-1)(y^2-y-1) = a^2$. Since we have $y(y-1) - (y^2-y-1) = 1$, $\gcd(y-1,y^2-y-1) = 1$. Thus $|y^2-y-1|=b^2$ for some $b \in \mathbb{Z}$. Suppose that $y > 2$. Then $(y-1)^2 < y^2-y-1 < y^2$, implying that there is no solution. The only values left to check are $y \in \{0, 1, 2\}$ It is easy to see that the set of all solutions is $(x,y) \in S = \{(0,0), (1,1), (0,2), (4,2)\}$

I believe this solution is correct.