Diophantine Here!

Upon observation, this Diophantine equation can be written as:

$y^3 = 4x^2 + 4x + 8 + 1$

which yields the following two factored forms:

$y^3 - 1 = 4(x^2 + x + 2) \Rightarrow (y-1)(y^2 + y + 1) = 4(x^2 + x + 2)$ (i);

$y^3 - 8 = 4x^2 + 4x + 1 \Rightarrow (y-2)(y^2 + 2y + 4) = (2x+1)^2$ (ii).

For expression (i), we now find that:

$y - 1 = 4; y^2 + y + 1 = x^2 + x + 2 \Rightarrow y = 5; 31 = x^2 + x + 2 \Rightarrow \boxed{y = 5; x = \frac{-1 \pm \sqrt{117}}{2}}$ ;

$y - 1 = 2; y^2 + y + 1 = 2(x^2 + x + 2) \Rightarrow y = 3; 13 = 2x^2 + 2x + 4 \Rightarrow \boxed{y = 3; x = \frac{-1 \pm \sqrt{19}}{2}}$ ;

$y - 1 = 1; y^2 + y + 1 = 4(x^2 + x + 2) \Rightarrow y = 2 ; 7 = 4x^2 + 4x + 8 \Rightarrow \boxed{y = 2; x = -\frac{1}{2}}$ ;

$y - 1 = 4(x^2 + x + 2); y^2 + y + 1 = 1 \Rightarrow \boxed{y = 0, -1}$ ;

$y - 1 = 2(x^2 + x + 2); y^2 + y + 1 = 2 \Rightarrow \boxed{y = \frac{-1 \pm \sqrt{5}}{2}}$ ;

$y - 1 = x^2 + x + 2; y^2 + y + 1 = 4 \Rightarrow \boxed{y = \frac{-1 \pm \sqrt{13}}{2}}$ .

For expression (ii), we now find that:

$y - 2 = 1; y^2 + 2y + 4 = (2x+1)^2 \Rightarrow y = 3; 19 = (2x+1)^2 \Rightarrow \boxed{y = 3; x = \frac{-1 \pm \sqrt{19}}{2}}$ ;

$y - 2 = 2x+ 1; y^2 + 2y + 4 = 2x + 1 \Rightarrow y^2 + y + 6 = 0 \Rightarrow \boxed{y = \frac{-1 \pm \sqrt{23}i}{2}}$ ;

$y - 2 = (2x=1)^2; y^2 + 2y + 4 = 1 \Rightarrow y^2 + 2y + 3 = 0 \Rightarrow \boxed{y = -1 \pm \sqrt{2}}$ .

Conclusion: there are NO positive integral pairs $(x,y)$ that satisfy the above Diophantine equation.