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.

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.    
