Solution:
The answers are $x=0$, $x=-1$, and $x=-2$, for all of which the product is $0=0^{2}$.

Assume that there is another solution. We can immediately rule out the case $x<-2$, as then the product is negative. So $x$ is a positive integer. Note that the greatest common divisor of $x$ and $x+1$ is $1$, as two consecutive integers cannot be multiples of the same prime. Likewise, $\gcd(x+2, x+1)=1$ and so $\gcd(x(x+2), x+1)=1$. Now $x(x+2)$ and $x+1$ are relatively prime positive integers whose product is a square, so they must both be squares (any prime dividing either of them must do so to an even power). But $x(x+2)=x^{2}+2x=(x+1)^{2}-1$, so now $x(x+2)$ and $(x+1)^{2}$ are two consecutive positive integers which are both squares, which is impossible (the difference between any two consecutive squares $a^{2}$ and $(a+1)^{2}$ is $2a+1 \geq 3$).