So I'm searching a certain $n\in \mathbb Z\,$ that satisfies $2^{n-1}\cdot n+1=y^2$.
How do I find the different solutions and how do I prove this?
I think that it would be possible to transform the equation to a sort of Pell equation: $$y^2-n \cdot 2^{n-1}= 1$$
Only this is not a Pell's equation because $n-1$ would have to be $2$.
Does another form exist?
Hint: Rewrite this equation as $2^{n-1} \cdot n =y^2-1=(y-1)(y+1)$. Then note that if $n>1$, the RHS is even, so $y$ is odd, and hence $y+1$ and $y-1$ are even. However, only one of them can be divisible by four, so either $y+1$ or $y-1$ is divisible by $2^{n-2}$. Now deduce that the RHS will get too large for larger $n$.