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?
 A: 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 $
$v_3(2^n+1) \ge v_3(n^2)$
$v_3(2+1)+v_(n) \ge v_3(n^2)$
$1+ \alpha \ge 2\alpha \implies \alpha =1,0$
$\implies v_3(n)=1,0$
$n=1 \text{ or } 3$
A: Andre's modification of a wrong answer :)
If $n=3^k$, then 
$$2^n+1=2^{3^k}+1=2^{3 \cdot 3^{k-1}}+1= (2^{3^{k-1}}+1)( 2^{2 \cdot 3^{k-1}}-2^{ \cdot 3^{k-1}}+1) $$
The second bracket is never divisible by $9$, thus by induction one can prove that $3^{2k-1}$ doesn't divide $2^n+1$.
Note: Since Geoff's answer was wrong, and this post doesn't make too much sense anymore, a simple observation:
If $n \neq 1$, then $3|n$.
Indeed let $p$ be the smallest prime divisor of $n$. 
Then $2^{p-1} \equiv 1 \mod p$ and $2^{2n} \equiv (-1)^2 \equiv 1 \mod p$.
Thus $2^d \equiv 1 \mod p$ where $d=gcd(p-1,2n)$. But no prime factor of $p-1$ can divide $n$, since $p$ is the smallest one, thus $gcd(p-1,n)=1$. Hence $d |2$.
$2^d \equiv 1 \mod p$ implies now that $p=3$.
This proves that $n=3^km$ for some $k \geq 1$ and $m $ relatively prime to $3$. I wonder if the first argument can be modified for this case:
$$2^n+1=2^{3^km}+1=2^{3 \cdot 3^{k-1}m}+1= (2^{3^{k-1}m}+1)( 2^{2 \cdot 3^{k-1}m}-2^{ \cdot 3^{k-1}m}+1) $$
Since $9$ doesn't divide the second bracket we get that $3^{2k-1}$ must divide $(2^{3^{k-1}m}+1)$ and repeating I think we get $3^{k}$ divides $2^m+1$...
It is easy to prove that $2^m \equiv -1 \mod 9$ implies $3 |m$ (this follows from $2^3 \equiv -1 mod 9$ and $ord(2)=6$).
Hence $k=1$, and we must have $n=3 m$ with $gcd(3,m)=1$...
Now, lets try the same again.
Suppose by contradiction $m \neq 1$. Let $q$ be the smallest prime factor of $m$.
Then 
$2^d \equiv 1 \mod q$ where $d=gcd(q-1,2n)$. But no prime factor of $p-1$ can divide $n$, excepting $3$,  Hence $d |6$.
This implies that 
$$2^6 \cong 1 \mod q \,.$$
Thus, the only possible values of $q$ is $q=7$.
But this is not possible since $2^{3m}+1 \equiv 1+1 \mod 7$, thus $7$ cannot divide $2^n+1$.
A: 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)
A: This was Problem 3 (first day) of the 1990 IMO. A full solution can be found here.
A: Since the links referenced in the question and answers are now mostly dead, I thought it would be worthwhile to cite the solution from the book The IMO Compendium, A Collection of Problems Suggested for The International Mathematical Olympiads: 1959 - 2009):

Let us assume $n>1$. Obviously $n$ is odd. Let $p \geq 3$ be the smallest prime divisor of $n$. In this case $(p-1,n)=1$. Since $2^n+1 \mid 2^{2n}-1$, we have that $p\mid 2^{2n}-1$. Thus it follows from Fermat's little theorem and elementary number theory that $p\mid (2^{2n}-1,2^{p-1}-1)=2^{(2n,p-1)}-1$.
  Since $(2n,p-1)\leq 2$, it follows that $p \mid 3$ and hence $p=3$.
Let us assume now that $n$ is of the form $n=3^kd$, where $2,3 \nmid d$. We first prove that $k=1$.
Lemma. If $2^m-1$ is divisible by $3^r$, then $m$ is divisible by $3^{r-1}$.
Proof. This is the lemma from (SL$97$-$14$) with $p=3, a=2^2, k=m, \alpha=1$, and $\beta=r$.
Since $3^{2k}$ divides $n^2 \mid 2^{2n}-1$, we can apply the lemma to $m=2n$ and $r=2k$ to conclude that $3^{2k-1}\mid n=3^kd$. Hence $k=1$.
Finally, let us assume $d>1$ and let $q$ be the smallest prime factor of $d$. Obviously $q$ is odd, $q\geq 5$, and $(n,q-1)\in \{1,3\}$. We then have $q \mid 2^{2n}-1$ and $q \mid 2^{q-1}-1$. Consequently, $q \mid 2^{(2n,q-1)}-1=2^{2(n,q-1)}-1$, which divides $2^6-1=63=3^2\cdot 7$, so we must have $q=7$. However, in that case we obtain $7 \mid n \mid 2^n+1$, which is a contradiction, since powers of two can only be congruent to $1,2$ and $4$ modulo $7$. It thus follows that $d=1$ and $n=3$. Hence $n>1 \Rightarrow n=3$.
It is easily verified that $n=1$ and $n=3$ are indeed solutions. Hence these are the only solutions. 

And here is the lemma referenced from SL$97$-$14$, again it can be found in the same book:

Lemma. Let $a,k$ be positive integers $p$ an odd prime. If $\alpha \geq 1$ and $\beta \geq 0$ are such that $p^\alpha \mid a-1$ and $p^\beta \mid k$, then $p^{\alpha + \beta} \mid a^k-1$.
Proof. We use induction on $\beta$. If $\beta = 0$, then $\frac{a^k-1}{a-1}=a^{k-1}+\dots+a+1\equiv k \pmod{p}$ (because $a \equiv 1$), and it is not divisible by $p$.
Suppose that the lemma is true for some $\beta \geq 0$, and let $k=p^{\beta+1}t$ where $p\nmid t$. By the induction hypothesis, $a^{k/p}=a^{p^\beta t}=mp^{\alpha+\beta}+1$ for some $m$ not divisible by $p$. Furthermore,
  \begin{align}
a^k-1&=(mp^{\alpha+\beta}+1)^p-1\\
&=(mp^{\alpha+\beta})^p+\dots+\binom{p}{2}(mp^{\alpha+\beta})^2+mp^{\alpha+\beta+1}.
\end{align}
  Since $p \mid \binom{p}{2}=\frac{p(p-1)}{2}$, all summands except for the last one are divisible by $p^{\alpha+\beta+2}$. Hence $p^{\alpha+\beta+1}\mid a^k-1$, completing the induction.

