Say for some positive integer $k$
$$ n2^n+1 = k^2 $$
This immediately suggests that factorisation might be a way out. But before we subject it to our number theoretic scrutiny, let us get our hands dirty and see where this leads.
| $n$ | $n 2^n$ | $n 2^n + 1$ |
|---|
| 1 | 2 | 3 |
| 2 | 8 | 9 ($\checkmark$) |
| 3 | 24 | 25 ($\checkmark$) |
| 4 | 64 | 65 |
| 5 | 160 | 161 |
| 6 | 384 | 385 |
It seems that there are no other solutions. Let us work on this angle. We will only be interested in the values of $n \ge 7.$
$$ \implies n 2^n = (k-1)(k+1) \tag{A} $$
Since $k+1$ and $k-1$ differ only by $2$, it suggests that not many powers of $2$ go into both. Say $k+1 =2 l \implies k-1 = 2(l-1).$ Therefore, equation $(A$) transforms into
$$ n 2^{n-2} = l(l-1) $$
Since gcd$(l,l-1)$ is $1,$ therefore, we can have one of the two cases
[(A)] $2^{n-2}$ divides $l \implies l \ge 2^{n-2}$ and $l-1 \le n \implies 2^{n-2} - n \le 1.$ But we see that $ 2^{n-2} - n = 25$ for $n=7$ and using induction we can prove that this is never less than or equal to $1$ for $n \ge 7.$ This part of the proof is left to the reader.
[(B)] Or, $2^{n-2}$ divides $l-1 \implies l-1 \ge 2^{n-2}$ and $ l \le n \implies 2^{n-2} - n \le -1, $ which is absurd for the same reasons as in the above part.
So, there is no value of $n \ge 7,$ that satisfies the condition. And the only values of $n$ that satisfy the conditions are $n=2,3.$
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