Answer is (a,b)=(2l,1),(l,2l),(8l^4-2l) forall l. Indeeed, assume b>1, k=a^2/2ab^2-b^3+1, expand we have x^2-2kb^2.x+k(b^3-1)=0. By vieta it has other root 2kb^2-a other from a. Now we have a claim: Any solution (a,b) either 2a=b or a>b. Indeed, Because b>1 and denominator>0 so 2a>=b thus a^2>=2ab^2-b^3+1 or a^2>=b^2(2a-b)+1>b^2 if 2a-b>0 so a>b. Assume exist (a1,b) and (a2,b) with b<>2a1,2a2, a1>a2 we have a1+a2=2kb^2, a1a2=k(b^3-1). Then a1>=kb^2 so a1a2>kb^3 thus the solution is described above

Seriously?

If $y=1 \implies x=2t \implies (x,y)=(2t,1)$ If $y=2x \implies OK \implies (x,y)=(t,2t)$ So, we only consider the pairs when $y\ge 2$ and $y\neq 2x$ Let $ S=\{(a,b) | (a,b) $satisfies the condition,$ b \ge 2$,$ b\neq 2a \}$ we choose $(a_0,b_0) \in S$ which $a_0$ is the minimum number between a of the elements. So, $a_0$ is the solution of the quadratic equation $x^2-2k{b_0}^2x+k{b_0}^3-k=0$. Let $e$ be the other root of the equation. $\implies a_0+e=2k{b_0}^2 $, $a_0e=k{b_0}^3-k$ We can easily find out that e is an positive integer. Claim: $b_0=2e$ If $b_0 \neq 2e \implies (e,b_0) \in S \implies e \ge a \implies e \ge k{b_0}^2$ Also, Since $2a_0{b_0}^2-{b_0}^3+1 \ge 1 \implies 2a_0 > b_0$ Since, $ {a_0}^2 \ge {b_0}^2(2a_0-b_0)+1 \implies a_0>b_0 \implies k{b_0}^3-k=a_0e > k{b_0}^3 \implies $ contradiction $\blacksquare$ We put $x=e$ and $b_0=2e$ at $x^2-2k{b_0}^2x+k{b_0}^3-k=0 \implies k=e^2 \implies a_0=8e^4-e$ So, if we consider $ S'=\{(a,b) | (a,b) $satisfies the condition, $(x,y) \neq (2t,1),(t,2t),(8t^4-t,2t) \}$, S' will be a null set. So our solutions are $(x,y)=(2t,1),(t,2t),(8t^4-t,2t)$

Ainnoway bluds copied a fricking IMO problem.