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: Beal's theorem gives min[math]\displaystyle{ (m, n) = 2 }[/math]. Squaring shows [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] where [math]\displaystyle{ \acute{y} = 4z }[/math] and [math]\displaystyle{ n }[/math] is odd. Odd [math]\displaystyle{ m }[/math] and [math]\displaystyle{ s \in {}^{\omega}\mathbb{N}^{*} }[/math] imply [math]\displaystyle{ x^m = \acute{y}\overset{\scriptsize{\grave{}}}{y} }[/math] where [math]\displaystyle{ s^m \ne \overset{\scriptsize{\grave{}}}{y} \in {}^{\omega}2\mathbb{N}^{*} }[/math] such that [math]\displaystyle{ x^m = 8\check{s}\acute{s} = 8.\square }[/math]