Theorem: 1 plus Square is not Perfect Power

The equation:
:$x^p = y^2 + 1$
has no solution in the integers for $x, y, p > 1$.

Proof:
Suppose $p$ is even. 
Write $p = 2 k$.
Then:
{{begin-eqn}}
{{eqn | l = 1
      | r = y^2 - x^{2 k}
}}
{{eqn | r = \paren {y - x^k} \paren {y + x^k}
      | c = Difference of Two Squares
}}
{{end-eqn}}
Since both $y - x^k$ and $y + x^k$ are integers, they must be equal to $\pm 1$.
Summing them up, we have $2 y$ is one of $-2, 0, 2$.
Thus $y$ is one of $-1, 0, 1$, and we ignore these solutions due to our condition $y > 1$.
Now suppose $p$ is odd.
Suppose $y$ is odd.
Then $x^p = y^2 + 1$ is even.
Hence $x$ is even.
Then:
{{begin-eqn}}
{{eqn | l = 0
      | o = \equiv
      | r = x^p
      | rr = \pmod 8
      | c = as $p \ge 3$
}}
{{eqn | o = \equiv
      | r = y^2 + 1
      | rr = \pmod 8
}}
{{eqn | o = \equiv
      | r = 1 + 1
      | rr = \pmod 8
      | c = Odd Square Modulo 8
}}
{{eqn | o = \equiv
      | r = 2
      | rr = \pmod 8
}}
{{end-eqn}}
which is a contradiction.
Hence $y$ must be even, and $x$ must be odd.
From Gaussian Integers form Euclidean Domain, we can define greatest common divisors on $\Z \sqbrk i$, and it admits unique factorization.
We factorize $y^2 + 1$:
:$x^p = y^2 + 1 = \paren {1 + i y} \paren {1 - i y}$
The greatest common divisors of $1 + i y$ and $1 - i y$ must divide their sum and product.
Their sum is $2$ while their product is $y^2 + 1$, which is odd.
Therefore we see that $1 + i y$ and $1 - i y$ are coprime.
From unique factorization we must have that both $1 + i y$ and $1 - i y$ is a product of a unit and a $p$th power.
By Units of Gaussian Integers, the units are $\pm 1$ and $\pm i$.
Hence
:$\exists u \in \set {\pm 1, \pm i}: \exists \alpha \in \Z \sqbrk i: 1 + i y = u \alpha^p, 1 - i y = \bar u \bar \alpha^p$
Since $p$ is odd:
:$1^p = 1$
:$\paren {-1}^p = -1$
:$i^p = \pm i$
:$\paren {-i}^p = -i^p = \mp i$
therefore there is some unit $u' \in \set {\pm 1, \pm i}$ such that $u'^p = u$.
By writing $\beta = u' \alpha$:
:$1 + i y = u'^p \alpha^p = \beta^p, 1 - i y = \bar \beta^p$
Write $\beta = a + i b$, where $a, b \in \Z$.
By Sum of Two Odd Powers:
:$2 a = \beta + \bar \beta \divides \beta^p + \bar \beta^p = 2$
this gives $a = \pm 1$.
We also have:
{{begin-eqn}}
{{eqn | l = 1 + y^2
      | r = \beta^p \bar \beta^p
}}
{{eqn | r = \paren {\beta \bar \beta}^p
}}
{{eqn | r = \paren {a^2 + b^2}^p
      | c = Product of Complex Number with Conjugate
}}
{{eqn | r = \paren {1 + b^2}^p
}}
{{end-eqn}}
since $1 + y^2$ is odd, $b$ must be even.
Hence:
{{begin-eqn}}
{{eqn | l = 1 + i y
      | r = \beta^p
}}
{{eqn | r = \paren {a + i b}^p
}}
{{eqn | r = \sum_{k \mathop = 0}^p \binom p k a^{p - k} \paren {i b}^k
      | c = Binomial Theorem
}}
{{eqn | o = \equiv
      | r = a^p + p a^{p - 1} i b 
      | rr = \pmod 4
      | c = $k \ge 2$ vanish as all terms containing $b^2$ is divisible by $4$
}}
{{end-eqn}}
In particular, comparing real parts gives $1 \equiv a^p \pmod 4$.
Since $p$ is odd, we have $a = 1$.
Now we have:
{{begin-eqn}}
{{eqn | l = 1 + i y
      | r = \paren {1 + i b}^p
}}
{{eqn | r = \sum_{k \mathop = 0}^p \binom p k 1^{p - k} \paren {i b}^k
      | c = Binomial Theorem
}}
{{eqn | ll = \leadsto
      | l = 1
      | r = \sum_{k \mathop = 0}^{\paren {p - 1} / 2} \binom p {2 k} b^{2 k} \paren {-1}^k
      | c = Comparing Real Parts; only even $k$ remain
}}
{{eqn | r = 1 - \binom p 2 b^2 + \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \binom p {2 k} b^{2 k} \paren {-1}^k
}}
{{eqn | r = 1 - \binom p 2 b^2 + \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac {p \paren {p - 1} } {2 k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k} \paren {-1}^k
      | c = Factors of Binomial Coefficient
}}
{{eqn | r = 1 - \binom p 2 b^2 + \binom p 2 b^2 \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k
      | c = Binomial Coefficient with Two
}}
{{eqn | ll = \leadsto
      | l = \binom p 2 b^2
      | r = \binom p 2 b^2 \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k
}}
{{eqn | ll = \leadsto
      | l = 1
      | r = \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k
}}
{{end-eqn}}
The summands on the {{RHS}} may not be an integer, but if we can show:
:In canonical form, the numerator of each summand is even
then the equation is never satisfied.
This is because the sum of all the terms will be a rational number with even numerator and odd denominator, which cannot equal to $1$.
Since $2 k + 1$ is always odd and $\paren {-1}^k \dbinom {p - 2} {2 k - 2}$ is always an integer, we only need to check $\dfrac {b^{2 k - 2} } k$.
Since $b$ is even:
:$2^{2 k - 2} \divides b^{2 k - 2}$
But we have:
{{begin-eqn}}
{{eqn | l = 2^{2 k - 2}
      | o = \ge
      | r = 2^k
      | c = as $k \ge 2$
}}
{{eqn | o = >
      | r = k
      | c = N less than M to the N
}}
{{end-eqn}}
Hence the largest power of $2$ that divides $k$ is less than $2^{2 k - 2}$.
Therefore the numerator of $\dfrac {b^{2 k - 2} } k$ is even.
And thus all the equations above are never satisfied.
So our original equation:
:$x^p = y^2 + 1$
has no solution in the integers for $x, y, p > 1$.
{{qed}}