Theorem: 1 plus Perfect Power is not Prime Power except for 9

The only solution to:
:$x^m = y^n + 1$
is:
:$\tuple {x, m, y, n} = \tuple {3, 2, 2, 3}$
for positive integers $x, y, m, n > 1$, and $x$ is a prime number.
This is a special case of Catalan's Conjecture.

Proof:
It suffices to show the result for prime values of $n$.
The case $n = 2$ is covered in 1 plus Square is not Perfect Power.
So we consider the cases where $n$ is an odd prime.
{{begin-eqn}}
{{eqn | l = x^m
      | r = y^n + 1
}}
{{eqn | r = y^n - \paren {-1}^n
      | c = as $n$ is odd
}}
{{eqn | r = \paren {y - \paren {-1} } \sum_{j \mathop = 0}^{n - 1} y^{n - j - 1} \paren {-1}^j
      | c = Difference of Two Powers
}}
{{eqn | r = \paren {y + 1} \paren {\map Q y \paren {y + 1} + R}
      | c = Division Theorem for Polynomial Forms over Field
}}
{{end-eqn}}
where $Q$ is a polynomial in one unknown and $R$ is a degree zero polynomial, so $R$ is a constant.
We have:
{{begin-eqn}}
{{eqn | l = R
      | r = \map Q {-1} \paren {-1 + 1} + R
}}
{{eqn | r = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^{n - j - 1} \paren {-1}^j
}}
{{eqn | r = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^{n - 1}
}}
{{eqn | r = \sum_{j \mathop = 0}^{n - 1} 1
      | c = as $n$ is odd
}}
{{eqn | r = n
}}
{{end-eqn}}
Hence we have $x^m = \paren {y + 1} \paren {\map Q y \paren {y + 1} + n}$.
Since $x$ is a prime, we have:
:$x \divides y + 1$
:$x \divides \map Q y \paren {y + 1} + n$
Hence $x \divides n$.
Since $x > 1$ and $n$ is a prime, we must have $x = n$.
Now we write $y + 1 = x^\alpha$.
Then we have:
{{begin-eqn}}
{{eqn | l = x^m
      | r = \paren {y + 1} \paren {\map Q y \paren {y + 1} + n}
}}
{{eqn | r = x^\alpha \paren {\map Q y x^\alpha + x}
}}
{{eqn | r = x^{\alpha + 1} \paren {\map Q y x^{\alpha - 1} + 1}
}}
{{end-eqn}}
For $\alpha > 1$, $x \nmid \map Q y x^{\alpha - 1} + 1$.
Hence $\alpha = 1$.
This gives $y + 1 = x = n$.
The equation now simplifies to:
:$\paren {y + 1}^m = y^n + 1$
Expanding:
{{begin-eqn}}
{{eqn | l = \paren {y + 1}^m
      | r = \sum_{j \mathop = 0}^m \binom m j y^j 1^{m - j}
      | c = Binomial Theorem
}}
{{eqn | r = 1 + \sum_{j \mathop = 1}^m \binom m j y^j
}}
{{eqn | r = y^n + 1
}}
{{eqn | ll= \leadsto
      | l = \sum_{j \mathop = 1}^m \binom m j y^{j - 1}
      | r = y^{n - 1}
}}
{{eqn | ll= \leadsto
      | l = \binom m 1 y^0
      | r = 0
      | rr= \pmod y
      | c = as $y > 1$
}}
{{eqn | ll= \leadsto
      | l = m
      | r = 0
      | rr= \pmod y
      | c = Binomial Coefficient with One
}}
{{end-eqn}}
hence we must have $y \divides m$.
By Absolute Value of Integer is not less than Divisors, $y \le m$.
Moreover, from $\ds \sum_{j \mathop = 1}^m \binom m j y^{j - 1} = y^{n - 1}$ we also have:
:$y^{n - 1} > \dbinom m m y^{m - 1} = y^{m - 1}$
Therefore we also have $n > m$.
This gives $y = n - 1 \ge m$.
The two inequalities forces $y = m$.
Now our original equation is further simplified to:
{{begin-eqn}}
{{eqn | l = \paren {y + 1}^y
      | r = y^{y + 1} + 1
}}
{{eqn | ll= \leadsto
      | l = \paren {1 + \frac 1 y}^y
      | r = y + \frac 1 {y^y}
      | c = dividing both sides by $y^y$
}}
{{end-eqn}}
From Real Sequence (1 + x over n)^n is Convergent:
:$\paren {1 + \dfrac 1 y}^y$ is increasing and has limit $e$.
Then we have for all $y \in \N$:
:$y + \dfrac 1 {y^y} < e < 3$
Since $\dfrac 1 {y^y} > 0$ and $y > 1$, we can only have $y = 2$.
This gives the solution $3^2 = 2^3 + 1$, and there are no others.
{{qed}}