Let $x,y$ be positive integers, such that $$3x(x^2+2)=y^2$$ since $$3\cdot 1(1^2+2)=3\times 3=9=3^2$$ $$3\cdot 2(2^2+2)=6\cdot 6=36=6^2$$ $$24\cdot 3(24^2+2)=72\cdot 578=204^2$$ so I have found three solutions $(x,y)=(1,3),(2, 6),(24, 204)$

Are there any other solutions?

ADD: In fact $$LHS=(x-1)^3+x^3+(x+1)^3$$