The question is : prove that $y^2 = x^3+(x+4)^2$ have no solutions in positive integers $x,y$.
I tried to play with the equation and get to $ x^3 = (y+x+4)(y-x-4)$, if $p|x$ then $p^3 | x^3$ and $ p | y+x+4$ or $ p | y-x-4$ , assume that $p$ divides both then we get that $ p | 2x-8$ and since $p|x$ we get that $ p | 8$ meaning $p=2$, so if we assume that $x$ is odd then $ p^3 | y+x+4$ or $p^3| y-x-4$ but not both and so we can write $y+x+4 = b^3 $ and $ y-x-4= a^3$, but that did not help me solving the problem.
Based on your argument, we can split into two cases:
In this case, as you said, we can assume that $y + x + 4 = b^3$ and $y - x - 4 = a^3$. Then we have $x = ab$. Hence, $b^3 - a^3 = 2x + 8 = 2ab + 8$. This means that $b > a$ and have the same parity, so $b - a \geq 2$.
However, this means that $2ab + 8 = b^3 - a^3 = (b-a)(b^2 + ba + a^2) \geq 2(b^2 + ba + a^2)$. Therefore, $a^2 + b^2 \leq 4$. A quick check show that this is impossible.
For the first case, $x = 2ab$ and $4ab + 8 = 2b^3 - 4a^3$, which implies $2ab + 4 = b^3 - 2a^3$, so $b = 2b'$ is even, which implies $2ab' + 2 = 4b'^3 - a^3$, which implies $a = 2a'$ is even, which implies $2a’b' + 1 = 2b'^3 - 4a'^3$, contradiction. The second case is similar.