For which $n \in \mathbf{N}^{*} n \times 2^{n+1}+1$ is a square?
Let $n$ be a solution, and $m$ such that $n \times 2^{n+1} + 1 = m^{2}$. Note that $n \times 2^{n+1}$ is even because $n+1 \geq 1$, so $m$ is odd. Let's write $m = 2k + 1$.

The equation then becomes $n \times 2^{n+1} = (m-1)(m+1)$, or equivalently $n \times 2^{n-1} = k(k+1)$. Among $k$ and $k+1$, one is even, and the other is odd. Suppose $k$ is even. Then $2^{n-1} \mid k(k+1)$, and by Gauss, $2^{n-1} \mid k$. Thus, $a 2^{n-1} = k$, and substituting into $n 2^{n-1} = k(k+1)$, we get $n = a(k+1)$, so $k+1 \mid n$. But then $k+1 \leq n$ and $2^{n-1} \leq k$.

Combining these, we get $2^{n-1} < 3$.

By induction, if we have an $n > 2$ such that $2^{n-1} > n$, then $2^n > 2n > n+1$, and this will still be true for $n+1$, which concludes the induction.

Thus, for $n \geq 3, 2^{n-1} > n$.

The case where $k$ is odd is resolved in the same way, by swapping $k$ and $k+1$. We still get an inequality that becomes false for large $n$, although it becomes false later than the one we have treated.

Finally, we get $3 \times 2^{3+1} + 1 = 49 = 7^2$ as the only solution.