Catalan's theorem
It holds that [math]\displaystyle{ \{(m, n, x, y) \in {}^{\omega}\mathbb{N}_{\ge 2}^4 : 1 + x^m = y^n\} = \{(3, 2, 2, 3)\} }[/math].
Indirect proof: Let [math]\displaystyle{ s, t \in {}^{\omega}\mathbb{N}^{*} }[/math]. Beal's theorem shows min[math]\displaystyle{ (m, n) = 2 }[/math]. Put [math]\displaystyle{ \acute{y} = 4z }[/math] and [math]\displaystyle{ \acute{n} \in {}^{\omega}2\mathbb{N}^{*} }[/math] to show by squaring [math]\displaystyle{ 1 + 4{\check{x}}^2 = (1 + \acute{y})^n }[/math] and [math]\displaystyle{ (1 + 2^rz)^n \equiv 1 + 2^r{\check{x}}^2 \equiv 1 }[/math] mod [math]\displaystyle{ 2^{\overset{\scriptsize{\grave{}}}{r}} }[/math] for every [math]\displaystyle{ r \in {}^{\omega}\mathbb{N}_{\ge 3} }[/math], and [math]\displaystyle{ 1 + x^2 \equiv 2 \ne 0 \equiv 2^n{\check{y}}^n }[/math] mod [math]\displaystyle{ 8 }[/math]. Putting [math]\displaystyle{ \acute{m} \in {}^{\omega}2\mathbb{N}_{\ge 4} }[/math] implies [math]\displaystyle{ 1 + x^m \ne y^2 }[/math] due to [math]\displaystyle{ x^m \ne s^m(2 + s^m) }[/math] and [math]\displaystyle{ t^m \ne 2 + s^m }[/math]. Thus [math]\displaystyle{ 1 + {8\check{x}}^3 = (1 + 2\check{y})^2 }[/math], [math]\displaystyle{ 2{\check{x}}^3 = s^3(1 + s^3) }[/math] and [math]\displaystyle{ s^3 - t^3 = t^3 - 1 }[/math] are left, and yield the solution specified.[math]\displaystyle{ \square }[/math]