The Diophantine equation $(1) \; y(x+y) = x^3-7x^2+11x-3$ is equivalent to $(2) \; 2y = -x \pm \sqrt{(x-2)(4x^2-19x+6)}$, implying $1 \leq x \leq 2$ or $x \geq 5$. Using (2) we obtain $y=-2$ or $y=1$ when $x=1$ and $y=-1$ when $x=2$. Next we consider (1) for $x \geq 5$. According to (2) the discriminant $(x-2)(4x^2-19x+6)$ have to be a perfect square, i.e. there exist two coprime positive integers $u$ and $v$ s.t. $x-2 = du^2$ and $4x^2-19x+6=dv^2$, where $d=GCD(x-2,4x^2-19x+6)$. The fact that $ 4x^2-19x+6 = 4(x-2)^2 - 3(x-2) - 16$ implies $d|16$. It also implies $4(du^2)^2 - 3du^2 - 16 = dv^2$, i.e. $(3) \; 4du^4 - 3u^2 - \frac{16}{d} = v^2$. Let us examine the following two cases: Case 1: $d=1$. By (3) $(4) \; 4u^4 - 3u^2 - 16 = v^2$, which means $2|u$ iff $2|v$. Hence, since $GCD(u,v)=1$, both $u$ and $v$ are odd. Futhermore, (4) is equivalent to $(8u^2-4v-3)(8u^2+4v-3) = 265$, which factors $8u^2 \pm 4v - 3 \equiv 1 \pmod{8}$. Combining this with the prime factorization of $265 = 5 \cdot 53$ and the fact that $53 \equiv 5 \pmod{8}$, the only possibility is $8u^2-4v-3=1$ and $8u^2+4v-3=265$, yielding $(8u^2-4v-3) + (8u^2+4v-3) = 1 + 265$, i.e. $u^2=17$. Clearly (1) has no solution in this case. Case 2: $d>1$. We know that $d|16$. If $d=16$, then by (3) $64u^4 - 3u^2 - 1 = v^2$, which has no solution since $(8u^2-1)^2 < 64u^4 - 3u^2 - 1 < (8u^2)^2$ for all positive integers $u$. Consequently $d \neq 16$, implying $d \in \{2,4,8\}$. In other words, ${\textstyle \frac{16}{d}}$ is even, which according to (3) and the fact that $GCD(u,v)=1$ means $u$ and $v$ are both odd. Again, by (3) and the fact that $2|d$, $\frac{16}{d} = 4du^4 - 3u^2 - v^2 \equiv 0 - 3 - 1\equiv 4 \pmod{8}$, implying $d=4$. Thus $16u^4 - 3u^2 - 4 = v^2$ according to (3), or alternatively $(32u^2-8v-3)(32u^2+8v-3) = 5 \cdot 53$. Using the fact that $32u^2 \pm 8v - 3 \equiv 5 \pmod{8}$, we see that the only possibility is $32u^2-8v-3=5$ and $32u^2+8v-3=53$, yielding $(32u^2-8v-3) + (32u^2+8v-3) = 5 + 53$, i.e. $u^2=1$. Therefore $u=1$, which means $x = du^2 + 2 = 4 \cdot 1^2 + 2 = 6$ and $y=-9$ or $y=3$ by (2). Conclusion: Equation (1) only have the five solutions $(x,y) = (1,-2), (1,1), (2,-1), (6,-9), (6,3)$.

It's nice problem. Here my solution: substitution, we have very useful reletion: \[ (2y+x)^2= (x-2)(4(x-2)^2-3(x-2)-16) \]. Let $x-2=t $, so $ t(4t^2-3t-16) $ is perfect square. $ t(4t^2-3t-16)\ge 0 $ $ \Longrightarrow $ $ t\in [-1,0]\cup [3, +\infty] $ If $ t\ge 3 $ , then $4t^2-3t-16\ge 0$ and we have $ gcd(t, 4t^2-3t-16) | 16 $ . hence, we have four cases: SOLUTION is $ (x,y)=(6, 3), (6, -9) $ ; If $t=0$, then $(x, y)=(2, -1); $ If $t=-1$, then $(x, y)=(1, -2), (1, 1);$ ANSWER: $ (x,y)=(6, 3), (6, -9), (2, -1), (1, -2), (1, 1); $