More integers

$2^{13} + 2^{10} + 2^{x} = y^{2}$

$\Leftrightarrow 2^{10}(2^{3}+1) + 2^{x} = y^{2}$

$\Leftrightarrow (2^{5}\times 3)^2 + 2^{x} = y^{2}$

$\Leftrightarrow 2^{x}= y^{2} - 96^2$

$\Leftrightarrow 2^{x}= (y + 96)(y-96)$

$\because y \in \mathbb{Z}^+, \textrm{ }y+96=2^m \textrm{ and } y-96=2^n, \left \{m,n \right \} \in \mathbb{Z}^+$

$2^m - 2^n = 192 = 2^6 \times 3$

$\Rightarrow 2^{m-6} - 2^{n-6} = 3$

$\Rightarrow m=8 \textrm{ and } n=6$

$\Rightarrow y=2^8-96=2^6+96=\boxed{160}.$

$2^{10}+ 2^{13}+2^x=y^2 \\ => (2^5)^2+ 2.2^5.2^7 +(2^7)^2 +2^x-(2^7)^2=y^2 \\ =(2^5+2^7)^2+(2^x-2^{14})=y^2$

Minimum ocurrs when $2^x-2^{14}=0 => 2^x=2^{14}$ then $y^2=(2^5+2^7)^2 => y=2^5+2^7=160$

If we take the square root of both sides to isolate $y$ and factor $2^{10}$ out of the left side, we obtain

$\sqrt {(2^{10})(1 + 2^3 + 2^x)} = y$

$\sqrt {(2^{10})(1 + 8 + 2^x)} = y$

$\sqrt {(2^{10})(9 + 2^x)} = y$

By using trial and error, we find that the lowest $x$ for which $y$ is an integer value is $4$ , and $y = 160$ .

Edit (as suggested by the asker):

We can see that the power of $2$ comes out of the square root $:$

$\sqrt{2^{10}} = 2^5$ .

$2^5\sqrt{9 + 2^x} = y$

For $y$ to be an integer, there must be an $x$ that satisfies $:$

$9 + 2^x = z^2, z \in \mathbb{Z}$

$x = 4$ is, in fact, the only solution, creating the Pythagorean triple $3, 4, 5:$

$9 + 16 = 25$

$z = 5 \in \mathbb{Z}$