We have $x^3 = (y-4)(y+4).$ From $\gcd(y-4,y+4)$ divides $(y+4)-(y-4)=8$ and $y+4\equiv y-4\pmod{8}$ it follows that either $\gcd(y-4,y+4)=1$ or $\gcd(y-4,y+4)=8.$ In either case both $y-4$ and $y+4$ must be cubes. The only cubes with the difference 8 are (-8,0) and (0,8), implying that $y=\pm 4$ and $x=0.$

$ (y-4)(y+4)=x^3$ $ y+4 >y-4$ So we take the perpuses: $ 1)$ $ y+4=x^3,y-4=1$(no integers solutions) $ 2)$ $ y+4=x^2,y-4=x$(no integers solutions) So there are no positives integers solutions...

No, dimitris, your reasoning only works for prime x. For example $ 6^3$ can also be factored as $ (2*3^2)*(2^2*3)$ or similar options.

maxal wrote: We have $x^3 = (y-4)(y+4).$ From $\gcd(y-4,y+4)$ divides $(y+4)-(y-4)=8$ and $y+4\equiv y-4\pmod{8}$ it follows that either $\gcd(y-4,y+4)=1$ or $\gcd(y-4,y+4)=8.$ Hello, maxal! If $y=8$, $\gcd(y-4,y+4)=4$ So we have to consider following cases. Case 1)$\gcd(y-4,y+4)=2$ $y-4=2k$, $y+4=2(k+4) (k:odd)$. Since $4k(k+4)=x^3$, $x=2m$.Then $k(k+4)=2m^3$ which is absurd. Case 2)$\gcd(y-4,y+4)=4$ $y-4=4k$, $y+4=4(k+2) (k:odd)$. Since $16k(k+2)=x^3$, $x=4m$.Then $k(k+2)=4m^3$ which is absurd. Thanks, Takeya.O.