Prime powers satisfying $2^n \pm 1$?

Question

Is there any solution to the equation

$$ p^k = 2^n \pm 1 $$

with prime $p > 3$ and natural number $k > 1$?

For $p=3$, is there any solution except the trivial case $3^2 = 2^3 + 1$?

What I did: Look into wikipedia and google on Mersenne primes (half of the $k=1$ case), and Prime powers, and went into the rabbit hole for both terms, but found nothing helping me out.

Links to dupes: (a) $2^n+1$ case, (b) $2^n-1$ case

Note: I'm asking the question on behalf of a retired family friend without internet, who came up with this question. I myself with 1yr of uni-level math ages ago were not able to find any prior work here nor was able to solve this question myself -- there must be thousands of such questions, and this may be trivial or very hard, I cannot judge properly.

Answer 1

Self-answer, but entirely thanks to @lulu:

The Catalan conjecture, now the theorem of Preda V. Mihăilescu, tells us that 8,9 are the only consecutive perfect powers.

The question looks for a $p^k$ and $2^n$ that differ by one, and the theorem linked above proves that for $k,n>1$ only the already mentioned case for $k=3$ is possible. Beyond that, only the trivial case $n=1$ remains, for which there is no solution $k>1$.

Answer 2

My answer from deleted question. I found the method at

Exponential Diophantine equation $7^y + 2 = 3^x$ by Gyumin Roh in 2015

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Let us start with the easy case. We believe the biggest integer solution to $ 3^u +1 = 2^v $ is $3+1 = 4,$ or $u=1,v=2.$ Begin with the $ 3^u +1 = 2^v $ and subtract $4$ from both sides, reaching $$3^u - 3 = 2^v - 4$$ with $u,v >0$ so $v \geq 2.$ Thus $$ 3(3^x -1) = 4 (2^y - 1) $$ where $x=u-1, y= v-2.$ We get equality if both $x,y$ are zero, and we think those are the biggest $x,y$ values that work.

Next, assume we have a solution with both $x,y > 0.$ This one is quick: since $4|(3^x - 1)$ we see that $x$ must be even. However, as soon as $x$ is even we find $3^x \equiv 1 \pmod 8$ and $8|(3^x - 1).$ This gives a contradiciton since $8 | 4(2^y - 1),$ then $2| 2^y - 1,$ contradicting the assumption that $y \neq 0$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

This case is more work:

We have $$ 3^u = 2^v + 1. $$ Subtract $9$ from both sides, we have $$ 3^u - 9 = 2^v - 8. $$ Taking $u = x + 2$ and $v = 3 + y$ gives $9 \cdot 3^x - 9 = 8 \cdot 2^y - 8,$ or $$ 9 (3^x - 1 ) = 8 (2^y - 1). $$ We think this is only possible with $x=y=0,$ so we assume there is a solution with $x,y > 0$ and get a contradiction.

First we have $9 |(2^y - 1),$ or $$ 2^y \equiv 1 \pmod 9.$$ This tells us that $ 6 | y. $ Meanwhile $$ 2^6 - 1 = 63 = 3^2 \cdot 7. $$ Therefore $7 | (3^x - 1).$ In turn, we find $6 | x.$

Next $$ 3^6 - 1 = 728 = 2^3 \cdot 7 \cdot 13. $$ Therefore $13 | (2^y - 1).$ In turn, we find $12 | y.$

Next $$ 2^{12} - 1 = 4095 = 3^2 \cdot 5 \cdot 7 \cdot 13. $$ Therefore $5 | (3^x - 1).$ In turn, we find $4 | x.$

Finally, $3^x - 1$ is divisible by $3^4 - 1 = 80.$ In particular, $3^x - 1$ is divisible by $2^4 = 16.$ However, this contradicts $2^y - 1 \neq 0$ and $y > 0.$ $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$