While solving this question on this site I arrived at the equation $$2^{2n}+2^m=k^2+k$$ where $m, n, k$ are natural numbers and $k\ge 2$. How to prove that $2^n=2^m=k$ if $m\le 2n$ ? I have found that $m=2n$ does not hold for the equation but was not able to do more. Your help is appreciated.
Thank you.
Factor it (assuming $2n \gt m$) as $2^m(2^{2n-m}+1)=k(k+1)$ The two factors on each side are relatively prime, so we must have $2^m=k, 2^{2n-m}+1=k+1$ or the other way around. If $k+1=2^m, k$ cannot be one more than a power of $2$ unless $k=3, m=2$