SOLUTION 2:
OK.We take 3 cases.
case1. Let, x=y. Then it can be easily told that no solution.
case2.Let, x>y. if x,y positive,then,$x^3$>$y^3$>$y^2$. so,$x^3$>$y^2$. & $x^3$-$y^2$>0.....(1). we get from equation, $x^3$-$y^2$=-7.contradiction. Now,if x,y negative(y may be positive),then here,$x^3$<$y^2$(as $y^2$ is always positive but $x^3$ is negative).then from equation,L.H.S. is positive & in R.H.S. $x^3$ is negative adding with 7 equals to L.H.S. which is positive that can easily be checked that no solution.
case3. Let, x<y.
Now, let $y=a(mod7)=>y^2=a^2(mod7)$ .let,$x=b(mod7)=>x^3=b^3(mod7)$.now,$a>0$,$b>0$ and $a<y$ ,$b<x$
Then, $y^2-x^3$=$[7(r)+a^2]-[7(p)+b^3]$>$a^2-b^3>0$
Now,
7|$y^2-x^3$
=>7|$a^2-b^3$
Now, $a^2-b^3=7$ [infinite descent!!!]
Here,Infinite descent occurs.So,
$contradiction!$
SO,There is no solution in integer.
[Proved]