find all integers $n$ such that $2^{n-1}*n+1$ is a perfect square. Clearly $n=0$
and i found that also $n=5$ gives a perfect square
And By representing the two functions , we found that there are only two solutions that are $n=0,5$
But I don't know how to prove that using elementary number theory.
 A: Suppose $2^{n-1} n + 1 = x^2$ with $n > 5$.  Then $2^{n-1} n = (x-1)(x+1)$ is a product of two integers that differ by $2$.  Corresponding to this we must have $x-1 = 2^k u$ and $x+1 = 2^{n-1-k} v$ where $uv = n$.  Now $\min(k,n-1-k) \le 2$, otherwise $x+1$ and $x-1$ would differ by at least $4$.
If $k \le 2$, $x+1 \ge 2^{n-3}$ and $x-1 \ge 2^{n-3}-2$ so
$$2^{n-1} n  =  (x+1)(x-1) \ge 2^{2n-6} - 2^{n-2} $$
which simplifies to 
$$ n + 1/2 \ge  2^{n-5}$$
and it is easy to see that this is false if $n \ge 9$.
Similarly if $n-1-k \le 2$, $ x-1 \ge 2^k \ge 2^{n-3}$ and
$x+1 \ge 2^{n-3}+2$ so
$$ 2^{n-1} n  \ge 2^{2n-6} + 2^{n-2}$$
and this is false if $n$ is too big.    
A: Robert Israel has already provided a nice answer, but it seems that you don't get it.
The idea in this answer is the same as the one in his answer. One can make his argument a bit simpler.
Let us prove that $2^{n-1}n+1$ is not a perfect square for $n\gt 5$.
Suppose that there exist positive integers $n,a$ such that $n\gt 5$ and
$$2^{n-1}n+1=a^2$$
Then, we have
$$2^{n-1}n=(a-1)(a+1)$$
Since LHS is even, RHS has to be even. Since $a-1$ and $a+1$ have the same parity, we see that both $a-1$ and $a+1$ have to be even. Therefore, there exist positive integers $p,q,r,s$ such that
$$a-1=2^pr,\qquad a+1=2^qs,\qquad p+q=n-1,\qquad rs=n$$ 
Suppose here that $p\ge 2$ and $q\ge 2$. Then, since both $a-1$ and $a+1$ are divisible by $4$, it follows that $2=(a+1)-(a-1)\ge 4$ which is impossible.
So, we have either $p=1$ or $q=1$.
If $p=1$, then $a+1=2^qs=2^{n-2}s\ge 2^{n-2}$ and $a-1\ge 2^{n-2}-2$ to have
$$2^{n-1}n\ge 2^{n-2}(2^{n-2}-2)\implies 2^{n-3}\le n+1$$
If $q=1$, then $a-1=2^pr=2^{n-2}r\ge 2^{n-2}$ and $a+1\ge 2^{n-2}+2$ to have
$$2^{n-1}n\ge 2^{n-2}(2^{n-2}+2)\implies 2^{n-3}\le n-1\implies 2^{n-3}\le n+1$$
So, in either case, it follows from the following lemma that $n\le 5$ which contradicts $n\gt 5$. $\quad\blacksquare$
Lemma : If $n$ is an integer such that $2^{n-3}\le n+1$, then $n\le 5$.
Proof for the lemma : 
It is sufficient to prove that if $n\ge 6\in\mathbb Z$, then $2^{n-3}\gt n+1$.
For $n=6$, we get $2^{n-3}=8\gt 7=n+1$.
Supposing that it holds for $n$ gives
$$2^{(n+1)-3}=2\cdot 2^{n-3}\gt 2(n+1)=(n+1)+1+n\gt (n+1)+1\qquad\square$$
