Proving that no positive integer solutions exist for $x^2 = y^{119}+1$ The $119$ looks very frightening so I factored $119=7\times 17$. And I think it is enough to prove the result for $x^2 = y^7+1$.
I then tried to move the $1$ around, and factoring. On one hand, $y^7 = (x+1)(x-1)$, and $\gcd(x+1,x-1) = 1 \text{ or } 2.$ For the first case, $x+1$ and $x-1$ are coprime, and their product is a seventh power, so they are both seventh powers, which is impossible since there difference is $2$. The other case is a bit difficult, so I retreated from this path.
On the other hand, $x^2 = (y+1)(y^6-y^5+y^4-y^3+y^2-y+1)$, and long division reveals that the $\gcd$ of the two factors can only be one of $1$ and $7$. I came up with a cubic fuction of $y$, and by case analysis on $y \mod 16$, I concluded that the sextic polynomial cannot be a square since it lies strictly between two consecutive squares. This closes the first case. But and for the second case both factors are $7$ times a perfect square. I did some unsuccessful attempts to reduce the whole thing to a Pell equation.
A moment of thought shows that $(x,y) = (1,0)$ is a solution (though not positive). And therefore any attempts to rule out solutions by inspecting the equation modulo some number is bound to meet failure. It might still work though, if we can somehow find a series of numbers $n_k$, and show that the solutions of $x^2\equiv y^7+1 \pmod {n_k}$ have some properties, so that they either result in $(1,0)$, or tends to infinity upon application of the Chinese Remainder Theorem, as $k$ tends to infinity. For small examples, I investigated the solutions modulo 29 and 43, since they are one more than a multiple of 7, and I can put some restrictions on the possible values of $y$. However I'm not able to generate an infinite number of the primes that suits my needs. And I'm a bit out of tools.

You have already found that $\gcd(x-1,x+1)=2$, from which it follows that
$$x-1=2^6a^7\qquad\text{ and }\qquad x+1=2b^7,$$
for some coprime $a,b\in\Bbb{Z}$ with $b$ odd, after changing the sign of $x$ if necessary. Then
$$b^{14}-(2a)^7=\left(\frac{x+1}{2}\right)^2-2(x-1)=\left(\frac{x-3}{2}\right)^2.\tag{1}$$
You have also found that the gcd of the two factors on the right hand side of
$$x^2=(y+1)(y^6-y^5+y^4-y^3+y^2-y+1),\tag{2}$$
is $7$. A similar argument shows that the left hand side of $(1)$ factors as
$$b^{14}-(2a)^7=(b^2-2a)(b^{12}-2ab^{10}+4a^2b^8-8a^3b^6+16a^4b^4-32a^5b^2-64a^6),\tag{3}$$
and that the gcd of these two factors divides $7$ because $b^2$ and $2a$ are coprime.
Claim: The two factors on the right hand side of $(3)$ are both divisible by $7$.
Proof. Suppose toward a contradiction that they are coprime. Equation $(1)$ shows that they are both perfect squares, so in particular $b^2-2a=c^2$ for some integer $c$. Because $y>0$ we have $a\neq0$ and so $c\neq b$, which implies
$$2|a|=|b^2-c^2|\geq2|b|-1,$$
and so$|a|\geq|b|$. On the other hand, because $x\geq0$ we have
$$|a|^7=\frac{|x-1|}{2^6}<\frac{|x+1|}{2}=|b|^7,$$
a contradiction. Hence we conclude that the two factors are not coprime.$\qquad\square$
Now it is easy to finish the proof; both factors on the right hand side of $(3)$ are divisible by $7$ and therefore their product $\big(\tfrac{x-3}{2}\big)^2$ is divisible by $7$, which shows that $x\equiv3\pmod{7}$. On the other hand you have already shown that both factors on the right hand side of $(2)$ are divisible by $7$, and therefore their product $x^2$ is divisible by $7$, which shows that $x\equiv0\pmod{7}$, a contradiction. This proves that there are no positive integers $x$ and $y$ such that $x^2=y^7+1$.

This answer is a  special case of Lemma 3.3 of the book Catalan's Conjecture by René Schoof (ISBN 978-1-84800-185-5). The proof of the claim is essentially the proof of this lemma as in the book.
