<s>[INST] 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?
 [/INST] 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 $2^n+1$.
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