Your idea matters-8

If y is odd, $y^2\equiv 1(mod 4) \Rightarrow x^3\equiv 2(mod 4)$ , Which is impossible. If y is even, $x^3+23\equiv 0(mod 4) \Rightarrow x^3\equiv 1(mod 4) \Rightarrow x\equiv 1(mod 4)$ . now, $y^2+4=x^3+27 \Rightarrow 4((\frac{y}{2})^2+1)=(x+3)(x^2-3x+9)$ but, $x\equiv 1(mod 4)\Rightarrow x^2-3x+9\equiv -1(mod 4)$ . So, $x^2-3x+9$ has at least a prime divisor p such that $p\equiv -1(mod 4)$ . Then, $p|(\frac{y}{2})^2+1$ bt we know,if q is a prime,a is a integer and $q|a^2+1$ , Then q=2 or $q\equiv 1(mod 4)$ . So contradiction!!!