Use geometrical approach

Denote by $k$ the ratio

$\huge\ \frac { { a }^{ 2 } }{ 2a{ b }^{ 2 } - { b }^{ 3 } + 1 }$ .

Because $k > 0$ we have $\large\ { 2a{ b }^{ 2 } - { b }^{ 3 } + 1 } > 0$ , so $\large\ a > \frac { b }{ 2 } - \frac { 1 }{ 2{ b }^{ 2 } }$ and hence

$\large\ a\ge \frac { b }{ 2 }$ . Using this and the fact that $k \ge\ 1$ , we deduce that $\large\ { a }^{ 2 }\ge { b }^{ 2 }\left( 2a-b \right) +1$ ,

So $\large\ { a }^{ 2 }>{ b }^{ 2 }\left( 2a - b \right) \ge 0$ . Hence either $a > b$ or $2a = b$ .

Now consider the two solutions $\large\ { a }_{ 1 }, { a }_{ 2 }$ to the quadratic equation

$\large\ { a }^{ 2 } - 2k{ b }^{ 2 }a + k({ b }^{ 3 } - 1) = 0,$

for fixed positive integers $k, b$ , and assume that one of them is an integer. Then other is also an integer because
$\large\ { a }_{ 1 } + { a }_{ 2 } = 2k{ b }^{ 2 }$ . We may assume that $\large\ { a }_{ 1 } \ge { a }_{ 2 }$ , and we have
$\large\ { a }_{ 1 } \ge k{ b }^{ 2 } > 0$ .

Furthermore, since $\large\ { a }_{ 1 }{ a }_{ 2 } = k\left( { b }^{ 3 } - 1 \right)$ , we obtain

$\huge\ 0 \le { a }_{ 2 } = \frac { k\left( { b }^{ 3 } - 1 \right) }{ { a }_{ 1 } } \le \frac { k\left( { b }^{ 3 } - 1 \right) }{ k{ b }^{ 2 } } < b.$

From the above it follows that either $\large\ {a}_{2} = 0$ or $\large\ 2{ a }_{ 2 } = b$ .

If $\large\ {a}_{2} = 0$ then $b^3 - 1 = 0$ , and hence ${a}_{1} = 2k, b = 1$

If $\large\ {a}_{2} = b/2,$ then $k = {b^2}/4,$ and $\large\ { a }_{ 1 }=\frac { { b }^{ 4 } }{ 2 } -\frac { b }{ 2 }.$

Hence , all satisfying pairs are $\huge\ \left( 2l,1, \right)$ or $\huge\ \left( l, 2l \right)$ or $\huge\ \left( 8{ l }^{ 4 }-l, 2l \right)$ for some positive integer $l$ .

Plugging $l = 1$ in one pair gives the least pair $(2, 1)$ ..

So, $x^y = 2$ .