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 implies [math]\displaystyle{ m = 3 }[/math] and [math]\displaystyle{ n = 2 }[/math], since vice versa for [math]\displaystyle{ \acute{x}, y \in {}^{\omega}2\mathbb{N} }[/math] the contradictions [math]\displaystyle{ 1 + x^2 \equiv 2 \equiv y^n \equiv 0 }[/math], [math]\displaystyle{ 1 + 4x^m \equiv 5 \equiv y^n \equiv 0 }[/math] and [math]\displaystyle{ 1 + 16x^m \equiv 1 \equiv y^n \equiv 0 }[/math] mod [math]\displaystyle{ 8 }[/math] arise. From [math]\displaystyle{ \acute{m} \in {}^{\omega}2\mathbb{N}_{\ge 4} }[/math], the contradiction [math]\displaystyle{ 1 + {\hat{x}}^m \equiv (1 + 2^{\acute{m}}s)^2 \equiv 1 + 2^ms + 4^{\acute{m}}s^2 \equiv y^2 }[/math] mod [math]\displaystyle{ 4^{\acute{m}} }[/math] follows. Thus [math]\displaystyle{ 1 + {\hat{x}}^3 = (\hat{y}^2 + 1)^2 }[/math], [math]\displaystyle{ 2{\hat{x}}^3 = s^3(s^3 + 1) }[/math] and [math]\displaystyle{ s^3 - t^3 = t^3 - 1 }[/math] for [math]\displaystyle{ t \in {}^{\omega}\mathbb{N}^{*} }[/math] are left and yield the solution specified.[math]\displaystyle{ \square }[/math]