Is it possible that $(x+2)(x+1)x=3(y+2)(y+1)y$ for positive integers $x$ and $y$? Let $x$ and $y$ be positive integers. Is it possible that $(x+2)(x+1)x=3(y+2)(y+1)y$?
I ran a program for $1\le{x,y}\le1\text{ }000\text{ }000$ and found no solution, so I believe there are none.
 A: It seems unlikely.  The search you have done seems to get us out of the law of small numbers.    Somebody's law says that when the reciprocal powers in an equation like this sum to less than $1$ you should expect finitely many solutions.  Here it is $\frac 23$.  The equation can be written $(x+1)((x+1)^2-1)=3(y+1)((y+1)^2-1)$ so you need $\frac {x+1}{y+1}$ to be very close to $\sqrt[3]3$  You could use the continued fraction to find convergents
A: The equation $x(x+1)(x+2) = 3y(y+1)(y+2)$ is equivalent to $\left(\frac{24}{3y-x+2}\right)^2 = \left(\frac{3y-9x-6}{3y-x+2}\right)^3-27\left(\frac{3y-9x-6}{3y-x+2}\right)+90$.
This is an elliptic curve of conductor $3888$. Cremona's table says its group of rational points is of rank $2$, and is generated by the obvious solutions $(x,y) \in \{-2;-1;0\}^2$
I am not sure how one would go about proving an upper bound for the integral solutions of the original equation. There are papers on the subject (for example, Stroeker and de Weger's "Solving elliptic diophantine equations: The general cubic case." seems to be applicable here)
Also, see How to compute rational or integer points on elliptic curves
