I am aware of Legendre's method for solving the Quadratic Diophantine Equation (QDE) of the form
$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \tag{1} $$
by transforming them into a Pell-type equation.
I have the following equation:
$$ x^3+4xy-8(n+1)y+4(n+1)^2=0 \tag{2} $$
where $n$ is a constant.
The leading monomial is of degree $3$. So it isn't a QDE.
Are there any methods for solving equations of type (2) with a cubic term?
The given equation is linear in $y$, so we can readily isolate it to get $$y=\frac{x^3+4(n+1)^2}{8(n+1)-4x}.$$ For $y$ to be an integer $x$ and $n+1$ must be even, say $x=2z$ and $n=2m-1$, and so $$y=\frac{z^3+2m^2}{2m-z}=\frac{2m^2(4m+1)}{2m-z}-(4m^2+2mz+z^2).$$ Given that $n$ is a constant, also $2m^2(4m+1)$ is a constant, and now it suffices to check all values of $z$ such that $2m-z$ divides $2m^2(4m+1)$.