This question is an exercise in chapter $5$ of Tate & Silverman's Rational Points on Elliptic curves.
Find all integer solutions to $x^2y + xy^2 = 240$.
Here is my work:
First we factor to find that \begin{align*} x^2y + y^2x &= 240 \\ xy(x + y) &= 240. \end{align*}
Now we list all possible factor pairs of $240$:
$$240\cdot1, 120\cdot 2, 4\cdot 60, 8\cdot 30, 16 \cdot 15, 3\cdot80, 6\cdot40, 12\cdot 20, 24\cdot 10, 48\cdot 5$$
We are looking for pairs in which one number can be written as $xy$ and another as $x + y$. By inspection, we find that the pairs $16\cdot 15, 12\cdot 20$ and $24\cdot 10$ are the only ones that work. So we have that
\begin{align*} 16 \cdot 15 &\implies x = 15,\: y = 1 \\ 12 \cdot 20 &\implies x = 10,\: y = 2 \\ 24 \cdot 10 &\implies x = 6,\: y = 4 \end{align*} which yields the solution set $\{(15, 1),\: (1, 15),\: (10, 2),\: (2, 10),\: (6, 4),\: (4, 6)\}$.
Is this method and solution set correct?
I don't think this approach is better for this problem. However, it is the case that whenever you have an eliptic curve and you have two rational solutions, then if the line through those points intesects the curve, it will give annother rational solution.
e.g. $(15,1)$ and $(10,2)$ are solutions. The line through those points is $x+5y = 20$
$(20-5y)(y)(20-5y+y) = 240\\ (5)(4-y)(y)(4)(5-y) = 240\\ (y)(4-y)(5-y) = 12\\ y^3 - 9y^2 + 20y -12 = 0$
Normally cubics are ugly, but you know two of the factors are $(y-1)$ and $(y-2)$ at which point finding the 3rd is easy.
$y^3 - 9y^2 + 20 y - 12 = (y-1)(y-2)(y-6)$
And plugging $y=6$ into the equation of the line gives $x=-10$
This technique helps to find rational solutions and not integer solutions, but often that is good enough.