The diophantine equation $y^3=x^2+4$ has only $4$ solutions. Is there an easy proof?

Question

Here

https://math.stackexchange.com/questions/2169360/how-to-determine-the-number-of-solutions-to-the-diophantine-equation#comment4461761_2169360

the equation $$y^3=x^2+4$$ is mentioned which has only the solutions $(\pm2,2)$ and $(\pm11,5)$. I suggested to subtract $8$ and factorize to get $$(y-2)(y^2+2y+4)=(x-2)(x+2)$$

Can I prove only with modular arithmetic , quadratic residues and similar methods, that there are no more solutions ? Does my approach help ?