Finding the solutions of an equation in positive intgers We are asked to find all positive integer solutions of the equation
$$x^7+7=y^2$$
or to show that it does not have any solutions. It is clear that $x$ can not be even. So $x$ is odd and $y$ is even (In fact $x$ is of the form $4k+1$).  Adding $121$ to both sides of the equation, we get $x^7+128=y^2+121$. Hence
$$(x+2)(x^6-2x^5+4x^4-8x^3+16x^2-32x+64)=y^2+121.$$ Since  the second expression on LHS is of the form $4k+3$, it does have a prime divisor $p$ with the same form. So
$$p|y^2+121$$, and from a previous lemma, we can conclude that $p|11$ and $p|y$, from which we obtain $p=11$ and $484| x^7+7$. If we proceed further, for $x^7+7$ to be divisible by 11, $x$ has to be of the form $11k+9$, where $k$ itself is shown to be of the form $11t+10$ giving us $x$ is of the form $121t+119$. How to continue in this way to show that such an equation does not have any solutions? Any help will be appreciated. Thank you.

I’m turning my comment into an answer.
Let $x$ be an integer such that $11|x^7+128$. Then clearly $11$ and $x$ are coprime and $11|(x^7)^3+128^3=x^{21}+2^{21}$. By (Pierre) Fermat’s little theorem, $11|x^{21}-x$, $11|2^{21}-2$ and thus $11|x+2$. Then write $x=11y-2$, then $\frac{x^7+128}{x+2}=\frac{(11y-2)^7+2^7}{11y}=\frac{11^2Ny+11\times 14y}{11y}=11N+14$ isn’t divisible by $11$.
In other words, for any integer $x$, $\frac{x^7+128}{x+2}$ is coprime to $11$. But we know from the OP that if $x^7+7=y^2$, then $11$ divides $\frac{x^7+128}{x+2}$, and we get a contradiction.
