Catalan's theorem
For [math]\displaystyle{ 1 + x^m = y^n }[/math] where [math]\displaystyle{ m, n, x, y \in {}^{\omega}\mathbb{N}_{\ge 2} }[/math], only the solution [math]\displaystyle{ m_0 = 3 }[/math], [math]\displaystyle{ n_0 = 2 }[/math], [math]\displaystyle{ x_0 = 2 }[/math] and [math]\displaystyle{ y_0 = 3 }[/math] exists.
Indirect proof: Assuming [math]\displaystyle{ \acute{x}, y \in {}^{\omega}2\mathbb{N} }[/math] yields the contradiction [math]\displaystyle{ 1 + x^m \equiv 2 \equiv y^n \equiv 0 }[/math] resp. [math]\displaystyle{ 4 }[/math] mod [math]\displaystyle{ 8 }[/math]. For [math]\displaystyle{ p \mid y,\ x = \hat{r},\ y = \hat{s} + 1 }[/math] and [math]\displaystyle{ 2 \lt m \ne n \gt 2 }[/math], subsequently [math]\displaystyle{ (\hat{s} + 1)^n - 1 \equiv \hat{r}^m \equiv -1 }[/math] mod [math]\displaystyle{ p \in {}^{\omega}\mathbb{P}_{\ge 3} }[/math] is assumed. Then also [math]\displaystyle{ (\hat{s} + 1)^{\hat{n}} - 1 = (\hat{s} + 2)^n\ \hat{s}^n \equiv \hat{r}^m \equiv -1 }[/math] mod [math]\displaystyle{ p }[/math] and [math]\displaystyle{ 2^{kn}(t^2 + t)^n\ 2^{-gn} \equiv 2^{lm}r^m \equiv -1 }[/math] mod [math]\displaystyle{ 2^{\text{max}(kn,\ lm)} }[/math] hold where [math]\displaystyle{ t = \check{s} }[/math] and maximal [math]\displaystyle{ g, k, l \in \mathbb{N}^{*} }[/math], from which [math]\displaystyle{ kn = lm }[/math] due to [math]\displaystyle{ (t^2 + t)2^{-gn}, r^m \in \mathbb{N} \setminus 2\mathbb{N} }[/math] and even [math]\displaystyle{ m = n }[/math] is implied contradicting the premise. Both [math]\displaystyle{ (\hat{s} + 1)^3 - 1 = \hat{r}^2 }[/math] are left, which can be refuted by the integer factorisation of [math]\displaystyle{ r }[/math] and [math]\displaystyle{ s }[/math], and [math]\displaystyle{ (\hat{s} + 1)^2 - 1 = 4s(s + 1) = \hat{r}^3 }[/math], which results in [math]\displaystyle{ r = s = 1 }[/math] and thus leads to the solution specified, too.[math]\displaystyle{ \square }[/math]