AIMO 2015 Q5

As the RHS is a perfect square, the LHS must also be a perfect square. Therefore,

$\begin{aligned} \color{#3D99F6}{2^{13}} + 2^{10} + 2^x & = y^2 \\ (2^5)^2 + \color{#3D99F6}{2(2^7)(2^5)} + \left(2^{\color{#D61F06}{\frac{x}{2}}}\right)^2 & = y^2 \quad \quad \small \color{#D61F06}{\Rightarrow \frac{x}{2} = 7} \\ \left(2^5 + 2^\color{#D61F06}{7}\right)^2 & = y^2 \\ \Rightarrow y & = 2^5 + 2^7 = 32 + 128 = \boxed{160} \end{aligned}$

You can also factorize them, for example:

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

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

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

Square root both sides and you'll get:

$y = 2^{5}\sqrt{9+2^{x-10}}$

The smallest possible value of x is 14 (to make $\sqrt{9+2^{x-10}}$ squarable), so try putting those numbers in and you'll find $\boxed{y = 160}$