Exponential diophantine equation involving a prime number

Solution 1:

As you noted: $$2^pn^p=(x+p^n)(x-p^n)$$

$$2^{p-2}n^p=\frac{x+p^n}2\frac{x-p^n}2$$ With the factors on the RHS being coprime and one of them odd and the other even.

If the first is the even one then $$\frac{x+p^n}2=2^{p-2}a^p$$ $$\frac{x-p^n}2=b^p$$ For integers $a,b$ with $ab=n$ because if a $k-$th power is the product of coprime factors, each of the factors is a $k-$th power. Then $$a^p=\frac{p^n+\sqrt{p^{p+1}+(p+1)^p}}{2^{p-1}}$$ $x\equiv\pm 1\pmod p$ so $a\equiv \pm 1$ and $$1<\left(\frac{p^n+\sqrt{p^{p+1}+(p+1)^p}}{2^{p-1}}\right)^{\frac 1p}=a<p-1.$$ For $p\geq 3$, and there's no number on this region that can satisfy $a\equiv \pm 1$. Contradiction.

If the second factor is the even one the proof is almost exactly the same $$\frac{x+p^n}2=a^p$$ $$\frac{x-p^n}2=2^{p-2}b^p$$ $$1<\left(\frac{p^n+\sqrt{p^{p+1}+(p+1)^p}}2\right)^{\frac 1p}=a<p-1$$ $$a\equiv x\equiv \pm 1\pmod p.$$ Contradiction.