18. 3. 43 ** Find all integer solutions \(x, y\) of the equation \(y^{2}=x^{3}+(x+1)^{2}\).
When $x=0$, the equation has two obvious solutions: $x=0, y= \pm 4$.
If $x(u-4)^{2} \geqslant(u-4)^{2}-y^{2}$; for $u>0$, from the original equation we get
$$
x^{3}=y^{2}-(x+4)^{2}=(y+x+4)(y-x-4) .
$$

Let $d=(y+x+4, y-x-4)=(y+x+4,2 x+8)$, then $(d, x)=(y+x+4, 2 x+8, x)=(y+4,8, x) \mid 8$. Since $d^{2} \mid x^{3}$, all prime factors of $d$ are prime factors of $x$, thus $d$ has no odd prime factors, $d=2^{k}$.
If $16 \mid d$, then $8 \| x$. At this time, $8 \| 2 x+8$, which contradicts $d \mid 2 x+8$.
If $d=1$, then $y+x+4=a^{3}, y-x-4=b^{3}, x=a b$. Thus $2 x+8=2 a b+8=a^{3}-b^{3}=(a-b)\left((a-b)^{2}+3 a b\right)$.

When $a-b=1$, $2 a b+8=1+3 a b, x=a b=7, y^{2}=7^{3}+11^{2}-464$, which is impossible (since 464 is not a perfect square).
When $a-b \geqslant 2$, $a^{3}-b^{3} \geqslant 2(4+3 a b)=8+6 a b>2 a b+8$.
Therefore, there is no solution in this case.
If $d=8$, then $y+x+4=8 a^{3}, y-x-4=8 b^{3}, x=4 a b .8 a b+8=2 x+8=8\left(a^{3}-b^{3}\right)=8(a-b)\left((a-b)^{2}+3 a b\right) \geqslant 8(1+3 a b)$, which is a contradiction. There is no solution in this case.

If $d=2$, then $x$ is even. Let $y+x+4=2 m, y-x-4=2 n,(m, n)=1$. Since $2 y=2(m+n)$, $m+n$ is even, thus both $m$ and $n$ are odd, which contradicts $x^{3}=4 m n$.

If $d=4$, then $x$ is a multiple of 4. Let $y+x+4=4 m, y-x-4=4 n,(m, n)=1$. From $x^{3}=4^{2} m n$ we get that one of $m$ and $n$ is a multiple of 4, and the other is odd. But $2 x+8=4(m-n)$, the left side is divisible by 8, while the right side is only divisible by 4, not by 8.
Therefore, the original equation has only two solutions $x=0, y= \pm 4$.