We have given the Diophantine equation $(1) \;\; 2^m p^2 + 1 = q^7$, where $p,q$ are prime numbers and $m$ in a natural number. First of all $q$ is odd by (1) since $m>0$. Futhermore (1) is equivalent to $(2) \;\; 2^m p^2 = q^7 - 1 = (q - 1)(q^6 + q^5 + q^4 + q^3 + q^2 + q + 1) = (q - 1)f(q)$. The fact that $q$ is odd implies $f(q) = \sum_{k=0}^6 q^k \equiv \sum_{k=0}^6 1^k = 7 \equiv 1 \pmod{2}$, which according to (2) means $2^m \mid q - 1$. Hence by (2) $(3) \;\; {\textstyle \frac{q - 1}{2^m} \cdot f(q) = p^2.}$ Now ${\textstyle \frac{q - 1}{2^m} < q - 1 < q + 1 < f(q)}$, which combined with (3) give us $q - 1 = 2^m$ and ${\textstyle p^2 = f(q) = \frac{q^7 - 1}{q - 1} = \frac{(2^m + 1)^7 - 1}{2^m} = 2^mk + 7,}$ yielding $2^m \mid p^2 - 7$. If $m \geq 2$, then $4 \mid p^2 - 7$, which is impossible. Hence $m=1$, which means $q = 2^m + 1 = 2^1 + 1 = 3$. Therefore ${\textstyle p^2 = \frac{q^7 - 1}{q - 1} = \frac{3^7 - 1}{3 - 1} = \frac{2187 - 1}{2} = \frac{2186}{2} = 1093,}$ yielding $p \not \in \mathbb{N}$. Conclusion:. Equation (1) has no solution.

Solar Plexsus wrote: $(3) \;\; {\textstyle \frac{q - 1}{2^m} \cdot f(q) = p^2.}$ Now ${\textstyle \frac{q - 1}{2^m} < q - 1 < q + 1 < f(q)}$, which combined with (3) give us $q - 1 = 2^m$ How do you get that $q - 1 = 2^m$ from these two equations??

We know that ${\textstyle \frac{q-1}{2^m}}$ is a positive integer. Futhermore ${\textstyle \frac{q-1}{2^m} \cdot f(q) = p^2}$, where $p$ is a prime, yielding ${\textstyle (\frac{q-1}{2^m},f(q)) \in \{(1,p^2), (p,p), (p^2,1)\}}$. Hence, since ${\textstyle \frac{q-1}{2^m} < f(q)}$, this implies ${\textstyle \frac{q-1}{2^m} = 1}$, i.e. $q - 1 = 2^m$.

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