So we have $2^k\mid n^2-5$. However by $\pmod{8}$ it follows that $2^k\in\{2,4\}$. Now we try cases and find only $\boxed{(k,n)=(0,4)}$

@above you missed $2^k=1$.

Ankoganit wrote: @above you missed $2^k=1$. Oops I read "positive" instead of "nonnegative"

Assume $n > 2$. Working modulu 8 we get $n^{2} =5 (mod 8)$ which is impossible. So the only solution is $(k,n) = (0,4)$.

Djaliloo wrote: Assume $n > 2$. Working modulu 8 we get $n^{2} =5 (mod 8)$ which is impossible. So the only solution is $(k,n) = (0,4)$. It should be $k > 2 $

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