Let $ 2^{n-1}n+1=N^{2}$, then if n>2, $ 2^{n-3}n=(N-1)/2.(N+1)/2=k(k+1)$. Now, we've got to make the LHS a product of two consecutive integers and since both of these integers should have the opposite pairity, we should make $ 2^{n-3}d$ and $ n/d$ consecutive integers, where d divides n But since $ 2^{n-3}>n+1$ holds for $ n>5$, we should have $ n<6$ Substituting for n values 1 through 5, we get that 5 is the only solution

the answer can only be 5..........lajanugen did it in a nice way

Let $2^{n-1}n+1=x^2$.Then $n\ge 3$ and $2^{n-1}n=(x+1)(x-1)$.Since $x+1,x-1$ are same parity, we can suppose that $x+1=2^as,x-1=2^bt$($a,b\ge 1$, s,t:odd, $a+b=n-1, st=n$).Since $x-1,x+1$ are distinct in mod 4,either $a$ or $b$ is $1$. Case1:a=1 $2=2(s-2^{n-3}t)\Leftrightarrow s-2^{n-3}t=1$. Then $t(2^{n-3}t+1)=n$, and $n\ge 2^{n-3}$.So $n\le 5$. Case2:b=1 $2=2(2^{n-3}s-t)\Leftrightarrow 2^{n-3}s-t=1$. Then $s(2^{n-3}s-1)=n$, and $n\ge 2^{n-3}-1$.So $n\le 5$. From case1,2, $n\le 5$.We can easily examine all $n\le 5$ cases, and we find only solution $\boxed{n=5}$.$\blacksquare$

Ans: $n=5, 2^{5-1}\cdot 5+1=81=9^2$. Proof: Suppose that we have $m\in \mathbb{N}$ such that \[2^{n-1}\cdot n+1=m^2 \iff 2^{n-1}\cdot n=(m-1)(m+1).\]Since we can easily see that $n=1$ yields no solution, we can just consider $n\geqslant 2,$ then the LHS is even which implies that $m$ must be odd. We now rewrite $m=2k+1$ for some $k\in \mathbb{N} \cup \{0\}.$ Then our equation rewrites into \[2^{n-3}\cdot n = k(k+1).\]We now split into two cases: Case 1: $k$ is even and $k+1$ is odd. This means that we can write $k=2^{n-3}\cdot a$ for some positive integer $a$ which is a divisor of $n$. So we have the following condition, \[2^{n-3}\cdot n=2^{n-3}\cdot a(2^{n-3}\cdot a+1)\geqslant 2^{n-3}(2^{n-3}+1)\]or $n\geqslant 2^{n-3}+1.$ However, it can be easily proven that $\forall n\geqslant 6,$ we have $2^{n-3}+1> n$ by a simple induction. So, we just have to check the cases when $n\leqslant 5.$ Case 2: $k+1$ is even and $k$ is odd. Similarly, we can write $k+1=2^{n-3}\cdot a$ for some positive integer $a$ which is a divisor of $n$. So we have the following condition, \[2^{n-3}\cdot n=2^{n-3}\cdot a(2^{n-3}\cdot a-1)\geqslant 2^{n-3}(2^{n-3}-1)\]or $n\geqslant 2^{n-3}-1.$ Again, we can also show that $\forall n\geqslant 6,$ we have $2^{n-3}-1> n$ by a simple induction. So, we just have to check the cases when $n\leqslant 5.$ Checking all the possible values for $n\leqslant 5,$ we find that $n=5$ is indeed the only solution.