Solve the Diophantine Equation $x^2 + 7 = y^5$. This is a duplicate question of Find integers solutions of $x^2+7=y^5$, however there was no full answer. The solutions $(\pm5, 2)$ and $(\pm 181, 8)$ have been found. 
The usual strategy for such a question is to work inside the ring of integers of $\mathbb{Q}(\sqrt{-7})$, which is $\mathcal{O} = \mathbb{Z}[ \frac{1+\sqrt{-7}}{2}]$. It turns out that this is a unique factorisation domain (which one can figure out by calculating its Class group). So it is natural to factor the equation as $(x - \sqrt{-7})(x+\sqrt{-7}) = y^5$. If we assume that $x-\sqrt{-7}$ and $x+\sqrt{-7}$ are coprime, we find that $x+\sqrt{-7} = \beta^5$ for a certain $\beta = a + b\frac{1+\sqrt{-7}}{2}\in \mathbb{Z}[\frac{1+\sqrt{-7}}{2}]$. Writing $c= 2a+b$ and expanding the fifth power, this gives the system of equations 
$$ c^5 -70 c^3 b^2 + 245 c^4 b = 32 x, $$
$$ 5 c^4 b -70 c^2 b^3 + 49 b^5 = 32. $$
Now with enough patience, one can show that this system has no solutions with $b \equiv c \pmod{2}$. 
However this contradicts the solutions that we have found. And indeed there's no reason for $x \pm \sqrt{-7}$ to be coprime when $x$ is odd. 
What is the approach to solve the remaining case of this diophantine equation?
One approach that I have tried is that the coprime condition holds inside the ring $\cal{O}[\frac{1}{2}]$. This gives the equation $x + \sqrt{-7} = (a+b\sqrt{-7})^5$ with $a,b \in \mathbb{Z}[\frac{1}{2}]$, which I am unable to solve.
 A: Consider the general case
$$x^2+7=y^m \tag{1}$$
(Integers $(x, y, m), \, m \geq 3)$
Let 
$$\rho = (1+\sqrt{-7})/2$$
Then as you are aware $(1, \rho)$ is a basis for the ring of integers of the field $\mathbb{Q}(-7)$. A standard factorization argument then compels us to devise a
$$\frac{x-1}{2}+\rho = \rho^{m-5}(U+\rho V)^m$$
For the case where $m=5$ one needs only consider the coefficients of $(U, V)$ and form
$$−U^5 − 15U^4V − 10U^3V^2 + 50U^2V^3 + 35UV^4 − 3V^5 = 1$$ 
This is a version of the Thue equation which I can solve using the R package NILDE.
Using this, the only solution is  the only solutions to the above equation are $(U, V ) = (−1, 0),(2, −1)$, which then give solutions proper as $(5, x, y)=(5, ±5, 2),(5, ±181, 8)$
Now, in Lesage, the author shows various partial results concerning the equation $(1)$, including the following.


*

*There are integer solutions to equation (1) with $m = 5, 7, 13$, this he proves by reducing to Thue equations, which he then solves by hand.

*There are no solutions to equation (1) for $m = 11$ and for $m$ prime
and $17 \leq m \leq 5000$. This he proves using classical algebraic number
theory, and a computational method.

*If $(x, y, m)$ is a solution to (1) then $m \leq 6.6 × 10^{15}$. This he proves
using lower bounds for linear forms in logarithms.

