For what values of n is $2^{2n} -1$ divisible by $4n+1$ For what values of n will the expresion $2^{2n} -1$ be divisible by $4n+1$.
I have checked using a computer and the values of 2n I get are 8,20,36,44,48,56,68,96,116,120,128,140,156,168,170,176 $\cdots$etc [For $2n<200$]
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I don't seem to find a relation between these values.I am curious to see if we can find a relation by applying simple number theory.

Writing $p:=4n+1$, your question gets restated as follows:
  $$ \text{For what values of $p$ is $\ 2^{\frac{p-1}2}\equiv 1\!\!\!\pmod p$?} $$
For $p$ prime, this means that $2$ is a quadratic residue mod $p$; that is, $p\equiv 1\pmod 8$. For $p$ composite, this implies $2^{p-1}\equiv 1\pmod p$; that is, $p$ is a base-$2$ Fermat pseudoprime. There are infinitely many Fermat's pseudoprimes, but they are rare. I doubt it is possible to characterize Fermat pseudoprimes $p$ with $2^{\frac{p-1}2}\equiv 1\pmod p$.
