Prove that the equation 2x−1=y2 has no integer solutions where x>1.

${\displaystyle y=2^x-1}$ has a root for ${\displaystyle x=0}$, because its asymptotic at ${\displaystyle y=-1}$, and is continually increasing. However, ${\displaystyle y^2=2^x-1}$ tells a different story. It still has a root at ${\displaystyle x=0}$, but its parabolic in nature in the horizontal direction.
The reason your function has no roots because of your domain of $x>1\cdots \frac{dy}{dx}>0,\text{for}x>1$ and at ${\displaystyle x=1,y=1}$. Hence, what can you conclude about the nature of the function and its position in a cartesian plane after ${\displaystyle x>1}$. Will it every hit the x-axis?

The point of a mathematical proof is not to be mathematical

Note that every square number is either $0\mathrm{mod}4\text{or}1\mathrm{mod}4$. It follows that for $x>2$, we have that $2^x-1=3\mathrm{mod}4$ and hence, cannot be a square. So if there is a solution it must happen when $x\le 2$. We can manually check that the only remaining potential solution of $x=2$ gives no integer solutions for y.