How many rationals of the form $\large \frac{2^n+1}{n^2}$ are integers?

This was Problem 3 (first day) of the 1990 IMO. A full solution can be found here.

How many rationals of the form $\large \frac{2^n+1}{n^2},$ $(n \in \mathbb{N} )$ are integers?

The possible values of $n$ that i am able to find is $n=1$ and $n=3$, so there are two solutions and this seems to be the answer to this problem.

But now we have to prove that no more of such $n$ exists, and thus the proof reduces to: Proving that $n^2$ does not divides $2^n+1$ for any $n \gt 3$.

Does anybody know how to prove this?

This was Problem 3 (first day) of the 1990 IMO. A full solution can be found here.

It's a bit late to post this answer. But I found this question can be solved using the Lifting the exponent lemma with so much ease.

Theorem: Let $x$ and $y$ be two integers and $n$ is an odd integer. Let $p$ be an od prime such that $p|x+y$ and none of $x$ and $y$ are divisible by $p$. Then we have,

$v_p(x^n+y^n)=v_p(x+y)+v_p(n)$

$v_p(N)$ denotes the highest power of $p$ which divides $N$.

Solution:

Claim: If $n$ divides $2^n+1$ then $n$ is a perfect power of $3$.

Proof:

Let $p$ be a smallest prime factor of $n$,

That means $2^n \equiv -1 \pmod p \implies 2^{2n} \equiv 1 \pmod p$. And also $2^{p-1} \equiv 1 \pmod p \implies 2^{\gcd(2n,p-1)} \equiv 1 \pmod p$

Now, since $p$ is the smallest divisor of $n$ the $\gcd(2n,p-1)=2 \implies 2^2 \equiv 1 \pmod p \implies p=3$, therefore, $n=3^m \cdot k \text { and } 3 \nmid k$, if $k$ is greater than $1$, the similar argument would show $3 |k$. Contradiction.

So we have $3^{\alpha} || n \implies 3^{\alpha+1} \nmid n $

$v_3(2^n+1) \ge v_3(n^2)$

$v_3(2+1)+v_(n) \ge v_3(n^2)$

$1+ \alpha \ge 2\alpha \implies \alpha =1,0$

$\implies v_3(n)=1,0$

$n=1 \text{ or } 3$

I've translated the handling of the problem into a certain notation, which I find very useful for formal algebraic manipulation of exponential diophantine problems, and provide a solution in that formalism. The Woeginger-solution was already linked by André Nicolas so my proposed way of solving this is now only for that reader who might be interested into that -hopefully: much general- formalism. Here is the link so far. (I'm a bit lazy to recode the text into latex/mathjax here after the original fiddling with word/word-to-pdf. Maybe I can put it into mathjax after the weekend, if there is any interest at all)