Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ A question asks

Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ for positive integers $x$ and $y$.

I tried factoring the LHS by adding $1$ to both sides so we get $(y+1)^3$ in the LHS. But I couldn't get any factorisation for the RHS, neither could think of any other ways to proceed.
How to proceed?
Thank you.

You can rewrite it as $$(y+1)^3 = x^3 + 5x^2 - 19x + 21$$
Now $y+1$ and $x$ differ for some integer, say $z$, so $(y+1) - x=z$, so we have $$(x+z)^3 =  x^3 + 5x^2 - 19x + 21$$ or $$ x^2(3z-5)+x(3z^2+19)+(z^3-21)=0$$ so we have a quadratic equation on $x$ with an integer parameter $z$: It will have an integer solution if it discriminat is a perfect square, so
$$(3z^2+19)^2-4(3z-5)(z^3-21) = d^2$$ or $$-3z^4+20z^3+114z^2+252z-59 =d^2$$
Notice that $$-3z^4+20z^3+114z^2+252z-59\geq 0$$ but that can not happend very often.
