Clearly, $n2^{n+1}+1$ is odd, so, if this number is a perfect square then
$n2^{n+1}+1=(2x+1)^2$, $x \in \mathbb{N}$, whence $n2^{n-1}=x(x+1)$.

The integers $x$ and $x+1$ are coprime,
so one of them must be divisible by $2^{n-1}$,
which means that the other must be at most
$n$. This shows that $2^{n-1} \le n+1$.

An easy induction shows that the above
inequality is false for all $n \ge 4$, and a direct
inspection confirms that the only
convenient values in the case $n \le 3$ are
$0$ and $3$.