For what $n$ is $\dfrac{p^n\cdot x-1}{p+1}$ integer, given $\lvert x\rvert_p=1$?
Where $\lvert x\rvert_p$ is inverse of the highest power of $p$ that divides $x$ (i.e. the p-adic absolute value)
I have for $p=2$ that:
Where $x\equiv3\pmod{6}$ there are no solutions.
Where $x\equiv 1\pmod{6}$, then $n\equiv 0\pmod{2}$.
Where $x\equiv 5\pmod{6}$, then $n\equiv 1\pmod{2}$.
For general $p$:
Where $x\equiv0\pmod{p}$ there are no solutions.
I can't see how complex the rest of the general form is likely to be, if attainable.
$$\frac{p^n\cdot x-1}{p+1} = \frac{p^n\cdot x-(-1)^n \cdot x}{p+1} + \frac{(-1)^n \cdot x-1}{p+1} = x \frac{p^n-(-1)^n}{p+1} + \frac{(-1)^n \cdot x-1}{p+1}$$
So it is determined by the parity of $n$.