Clearly $n > 2$. $n2^{n-1} = (x-1)(x+1)$ hence there exists an odd divisor $p$ of $n$ such than $x - 1 = 2p$ and $x+1 = 2^{n-2}\frac{n}{p}$. Thus we get $p(p+1) = 2^{n-3}n$. Since $p$ is odd we have $2^{n-3} \mid p+1$ and consequently $p \geq 2^{n-3} - 1$. Hence $2^{n-3}n = p(p+1) \geq 2^{n-3}(2^{n-3}-1)$ equivalently $n \geq 2^{n-3} - 1$. This implies $n < 6$ and by checking all the options we get that $n=5$ is the only solution.

$\color{blue} \boxed{\textbf{SOLUTION}}$ $n2^n-1=a^2 \implies n 2^{n-1}=(a-1)(a+1)$ $(a-1),(a+1)$ both are even as they differ by $2$ and equal to $n2^{n-1}, n \geq 2$ Therefore, there exist positive integers $p,q,r,s$ $a-1=2^{p}r, a+1=2^{q}s, p+q=n-1, rs=n$ $\color{red} \textbf{Case 1 :}$ $p,q \geq 2$ $\implies 4|(a+1), 4|(a-1)$ But it gives $2=(a+1)-(a-1) \geq 4$ So, $p,q \ngeq 2$ $\color{red} \textbf{Case 2 :}$ $p=1 \implies a+1=2^q s=2^{n-2} s \geq 2^{n-2}$ $a-1 \geq 2^{n-2}-2$ $n2^{n-1} \geq 2^{n-2} (2^{n-2}-2) \implies n+1 \geq 2^{n-3}$ $\color{red} \textbf{Case 3 :}$ $q=1 \implies (a-1)=2^{p}r=2^{n-2} \geq 2^{n-2}$ And $(a+1) \geq 2^{n-2}+2$ $n2^{n-1} \geq 2^{n-2} (2n-2+2) \implies n-1 \geq 2^{n-3} \implies n+1 \geq 2^{n-3}$ So, $\boxed{n+1 \geq 2^{n-3}}$ It can be proved by Induction. $\textbf{proof}$ Assume, $n \geq 6$ for $n=6, 6+1 \ngeq 2^{6-3}$ Let, for $n, 2^{n-3} \ge n+1$ So, $2^{(n+1)-3}=2.2^{n-3} \ge 2(n+1) = n+1+n+1 \ge (n+1)+1$ $\blacksquare$ Therefore $n \leq 5$ Plugging values we get $\color{red} \boxed {\textbf{n=5}}$

Clearly $n \geq 3$, so, $n \cdot 2^{n-1} + 1$ is a perfect square odd, then $n \cdot 2^{n-1} + 1 = (2k + 1)^2$ $= 4k(k+1) + 1$ $\Longrightarrow n \cdot 2^{n-3} = k(k+1)$, now take $n = 2^a \cdot b$ ($b$ is odd) thus $n \cdot 2^{n-3} = 2^{2^{a} \cdot b + a - 3}$ $\cdot b = k(k+1)$. Now we have two cases. $i)$ $\begin{cases} k = b \\ k + 1 = 2^{2^{a} \cdot b + a - 3} \end{cases}$ $\Longrightarrow k + 1 = 2^{2^a \cdot k + a - 3} \geq 2^{k - 3}$ but $2^{k-3} > k + 1$ $\forall k \geq 6$ (induction) and checking $k \leq 5$, we've no solution. $ii)$ $\begin{cases} k + 1 = b \\ k = 2^{2^{a} \cdot b + a - 3} \end{cases}$ $\Longrightarrow k = 2^{2^{a} \cdot k + 2^a + a - 3} \geq 2^{k-2}$ but $2^{k-2} > k$ $\forall k \geq 5$ (induction) and checking $k \leq 4$ we have $k = 4$ and $a = 0$ $\Longrightarrow b = 5 \Longrightarrow n = 5$. $\therefore n = 5$ is the only solution.

Let $n\cdot2^{n-1}+1=x^2.$ Thus, we realize that $x$ is odd $\Longrightarrow$ $x=2k+1$. Substituting, we have $n\cdot2^{n-1}+1=4k^2+4k+1$ $\Longrightarrow$ $n\cdot2^{n-3}=k^2+k=k(k+1)$. So let's divide it into two cases: First case: $k$ being odd. In that case, $2^{n-3}|k(k+1)$ $\Longrightarrow$ $2^{n-3}|k+1$ $\Rightarrow$ $k+1\ge2^{n-3}.$ We then have to $n\cdot2^{n-3}=k(k+1)\ge2^{n-3}\cdot(2^{n-3}-1)$ $\Longrightarrow$ $n\ge2^{n-3}-1$. It is easy to prove by induction that $6\ge{n}$ and, testing, we only found $n=5$. Second case: $k$ being even. This case is similar to the previous one. We found $n\ge2^{n-3}+1$ $\Longrightarrow$ $6\ge{n}$. Again, we only found $n=5$. The only solution to the problem is $n=5$.

$n*2^{n-1}+1 = k^2$ $n*2^{n-1} = k^2-1$ if k even then $n=1$ but $n=1$ not a sols so $n\geq 2$ $n*2^{n-1} = (k-1)(k+1)$ $k^2 \equiv 1 \mod 2$ thus $k \equiv 1 \mod 2$ let $k = 2m-1$ $n*2^{n-1} = (2m-2)*2m$ $n*2^{n-3} = m(m-1)$ this lead to whether $2^{n-3} \mid (m,m-1)$ since $\gcd(m,m-1) = 1$ it divide only one of them however either cases show $2^{n-3}-1 \leq m$ $n*2^{n-3} \geq (2^{n-3}-1)(2^{n-3}-2)>2^{n-3}-3*2^{n-3}$ $n*2^{n-3} >2^{2(n-3)}-3*2^{n-3} $ $n>2^{n-3}-3$ since exponent run faster than linear so just check small cases $n=2,3,4,5,6,....$ $n=2 $ true $n=3 $ true $n=4 $true $n=5 $ true $n=6 $ true now for $n >6$ assume it is still true $n>2^{n-3}-3$ sub $n = k+6$ where $k \geq 1$ $k+9>2^{k+3}$ $\frac{k+9}{8}>2^k >k$ contradiction so just check the few cases I mentioned above $n=2 , 2*2^1+1 =5$ $n=3 , 3*2^2+1 = 13$ $n=4 , 4*2^3+1 = 33$ $n=5 ,5*2^4+1 = 81$ $n=6 , 6*2^5+1 1=193$ it easy to see only n=5 satisfy

The answer is only $n = 5$. It's easy to check all values $n \leq 5$, so henceforth assume $n \geq 6$. Suppose $n$ works and write $n \cdot 2^{n - 1} + 1 = k^2$ for some positive integer $k$. Rearranging gives \[(k + 1)(k - 1) = n \cdot 2^{n - 1},\]and since one of $k + 1$ and $k - 1$ is $2$ mod $4$, it's at most $2n$. Thus $k - 1 \leq 2n$, so \[(k + 1)(k - 1) \leq 2n(2n + 2) = 4n(n + 1).\]Putting this together with the above equation and checking cases yields $n \leq 5$, a contradiction. $\square$

Let $n\cdot 2^{n-1}+1=x^2$ If $p$ is a prime number such that $p\mid n$, then $x-1=2p$ and $x+1=2^{n-2}\cdot \frac{n}{p}$, therefore, $p(p+1)=n2^{n-3}\implies 2^{n-3}\mid p+1\implies 2^{n-3}-1\leq p \leq n\implies n\geq 2^{n-3}-1\implies n<6$, then by checking the options we have that $\boxed{n=5}$ is the only solution.

Assume $n \geq 6$. We have $n\cdot2^{n - 1} + 1 = k^2 \implies (k - 1)(k + 1) = n\cdot2^{n - 1}$. Clearly, $k$ is odd. If $k \equiv 1\mod{4}$, we have $v_2(k - 1) = n - 2 \implies k \geq 2^{n - 2} + 1$, and $(k - 1)(k + 1) \geq 2^{n - 2} \cdot (2^{n - 2} + 2) = 2^{n - 1}(2^{n - 3} + 1) > 2^{n - 1} \cdot n $, since $2^{n - 3} + 1 > n$ for $n \geq 6$, so a contradiction here. If $k \equiv 3\mod{4}$, we have that $v_2(k + 1) = 2^{n - 2} \implies k \geq 2^{n - 2} - 1 \implies (k + 1)(k - 1) \geq 2^{n - 1}(2^{n - 3} - 1) > 2^{n - 1}\cdot n$, since $2^{n - 3} - 1 > n$ for $n \geq 6$, a contradiction So, checking $n = 1, \dots, 6$ gives $n = 5$ as the only solution.