Show that $x^3-x^2+8=y^2$ has no integer solution
The factorization in the comment by Amin235 is the key. If $x$ is odd, then the term $x^2-2x+4$ is always $-1$ modulo $4$ (and hence has a prime divisor with the same property, which must divide $x^2+y^2$, a contradiction). If $x$ is even, then it is necessarily $2$ modulo $4$ (since otherwise $y^2$ is divisible by $8$, but not by $16$), whereby the same is true of $y$. It follows that $x^2+y^2$ is $8$ modulo $16$, while $(x+2)(x^2-2x+4)$ is $0$ modulo $16$.