we shall consider two cases.                                                           case 1: let n=2m then 2^2m +65=k²    ;65=(k-2^m)(k+2^m)            5*13=(k-2^m)(k+2^m) both factors on LHS are positive integers, second k+2^m greater than first k-2^m . taking k-2^m =5, k+2^m =13 2^m+1 = 8, m=2, n=4 taking k-2^m =1 ;k+2^m =64 2^m+1 =64, m=5, n=10

Now, we deal with the second case if n=2m+1                                2*2^2m +65=k² k² -2*(2^m )²= 65 since RHS=0 (mod 5) LHS must be 0 (mod 5) first of all, if k=0 (mod 5) then 2^m must be 0 (mod 5), a contradiction we can easily see that only 1 and -1 are quadratic residues mod 5 but 2*(2^m )² is only 2 or -2 mod 5 so there are no solutions in this case so n=4, 10 are the only solutions.

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