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 right hand side 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}}