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We need to find all integers $x>1$ such that $$\frac{2^x+1}{x^2}$$is an integer. I have pointed out that $x=3$ works. And none of the even integer will work. How do I proceed?

User8976
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    This is a number theory problem, so I think you should change your tags to be more likely to get help. How much number theory do you know? Enough to show that 3 is the only prime value of x that works? – Barry Smith Jun 25 '20 at 01:03
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    if $x$ is prime, $x$ divides $2^x-2$, but this implies $x$ divides $2^x+1$, so $x$ divides $3$ – J. W. Tanner Jun 25 '20 at 01:24
  • A partial result: if $x$ divides $2^x+1$ (never mind $x^2$), $x$ must be a power of $3$ times a finite number of other values, ${1,171,13203,2354697,10970073,22032887841}$. Of these, you might be able to show directly that $x^2 \nmid 2^x+1$.

    https://www.maths.ed.ac.uk/~chris/papers/n_divides_2to_nplus1.pdf

     – Integrand Jun 25 '20 at 02:42
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    Does this answer your question ?https://math.stackexchange.com/questions/203976/find-all-n1-such-that-dfrac2n1n2-is-an-integer – ShBh Jun 25 '20 at 02:57
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    Does this answer your question? Find all $n>1$ such that $\dfrac{2^n+1}{n^2}$ is an integer. – jijijojo Jun 25 '20 at 05:34

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