Junior Balkan Mathematical Olympiad

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

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 $3$.