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.

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
