Prove that $n^7+7$ can never be a perfect square.

To show that $n^7 + 7 = x^2$ has no solution we shall use the fact that

$n^7 - 7 = x^2$ has one solution namely $n = 2$ and $x = 11$.

Suppose $n^7 + 7 = x^2$. Adding to $2^7 - 7 = 11^2$ we get $ n^7 + 2^7 = x^2 + 11^2 $ Call $m = n + 2$. Then

$ n^7 + 2^7 = (m - 2)^7 + 2^7 = $

$$\sum_{0 \le k \le 7} \binom{7}{k} m^k (-2)^{7-k} + 2^7$$

$$ = \sum_{1 \le k \le 7} \binom{7}{k}(-1)^{k-1} m^k (2)^{7-k} $$

$m$ can be factored out and you get

$$n^7 + 2^7 = m \sum_{1 \le k \le 7} \binom{7}{k} (-1)^{k-1} m^{k-1} (2)^{7-k}$$

Let's write $n^7 + 2^7 = m\cdot M.$

Note that $M = m^6 -7\cdot 2\cdot m^5 + - + -21\cdot 2^5m + 7\cdot 2^6$

Hence $\gcd(m,M)$ is a divisor of $7\cdot 2^6$

Note that in $\Bbb{Z}_4$, the squares are $0$ and $1$. Hence $n^7 + 7 = 0$ or $1$ in $\Bbb{Z}_4$, which implies that $n$ is odd.

This implies that $m$ is odd. Hence $\gcd(m,M)$ is either $1$ or $7$.

Observe that in $\Bbb{Z}_7$, the squares are $0,1,2,4$. Hence the sum of $2$ squares can be a multiple of $7$ only if both squares are themselves multiples of $7$. Since $11$ is not, $x^2 + 11^2$ can't be a multiple of $7$.

Therefore $\gcd(m,M)=1$

Now if $m \cdot M$ is a sum of $2$ squares, its square free part has prime factors only of the form $p = 4k + 1$.

Hence the same is true for m, since $\gcd(m,M) = 1$.

This implies that $m = 1 \pmod 4$ and $n = m - 2 = - 1 . \pmod 4$

Hence $n^7 + 7 = -1 -1 = 2 \pmod 4$ and is not a square.

This yields a contradiction: $n^7 + 7$ can't be a square.

Here is an alternative proof that is hopefully slightly simpler.

We will make use of a simple theorem whose proof is skipped here: For any odd integer number $m$, we have $m^2\equiv 1(mod 4)$.

Assume on the contrary, $n^7+7$ is a perfect square, then there exists an integer $x$ such that $n^7+7=x^2$.

If $x$ is odd, $n$ would have to be even. And the left side would be $3 (mod 4)$ that contradicts the fact that the right side is $1 (mod 4)$ by the theorem above. So $x$ has to be even, and $n$ has to be odd. Further more, applying the above theorem on $n^6=(n^2)^3$, we know $n^7\equiv n (mod 4)$. Note the right hand size is $0 (mod 4)$ since $x$ is even, we must have $$n\equiv 1 (mod 4).$$

Let's denote $x=2b$ where $b$ is another positive integer. Now we get $n^7+7=4b^2$. Further utilizing the fact that $7=2^7-11^2$. We can rewrite the above equation as $n^7+2^7=4b^2+11^2$, or $$(n+2)(n^6-2n^5+4n^4-8n^3+16n^2-32n+64)=4b^2+11^2.$$

Now, notice $n+2\equiv 3 (mod 4)$, it MUST have a prime factor $p$ with $p \equiv 3 (mod 4)$, and $p \mid (4b^2+11^2)$. We next consider two cases:

Case one: $11 \nmid b.$

Since $p \mid (4b^2+11^2)$, we know $p \neq 11$, $p \nmid b$, $4b^2 \equiv -11^2(mod p)$. Note $\frac{p-1}{2}$ is odd, ${\left(4b^2\right)}^{\frac{p-1}{2}} \equiv {\left(-11^2\right)}^{\frac{p-1}{2}}(mod p)$, ${(2b)}^{p-1}=-11^{p-1}(mod p)$. Applying Fermat's little theorem, we get $1 \equiv -1 (mod p)$, a contradiction.

Case two: $11 \mid b.$

Let $b=11 \times a$, $a$ is an integer. We have $$ (n+2)(n^6-2n^5+4n^4-8n^3+16n^2-32n+64)=(4a^2+ 1)\times 11^2. $$ We can manually verify that $$ 11 \nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $$ for $n \equiv -5 (mod 11)$ to $n \equiv 5 (mod 11).$ Hence $11^2 \mid n+2.$ Note $11^2 \equiv 1(mod 4)$, we still have the claim that $\frac{n+2}{11^2}$ has a prime factor $p$ with $p \equiv 3(mod 4)$, and $p \mid 4a^2+1$. $$ 4a^2 \equiv -1 (mod p)\\ {\left(4a^2\right)}^{\frac{p-1}{2}} \equiv -1 (mod p) \\ {\left(2a\right)}^{p-1} \equiv -1 (mode p) \\ 1 \equiv -1 (mod p),$$ a contradiction.