2 is Power-"Full"!

$\begin{aligned} \left(2^4 + 2^u \right)^2 & = 2^8 + 2\dot{}2^{4+u} + 2^{2u} \\ \text{For } \quad 2^8 + 2^{15} + 2^{x} & = 2^8 + 2\dot{}2^{4+u} + 2^{2u} \\ \Rightarrow & \begin{cases} 14 = 4 + u & \Rightarrow u = 10 \\ x = 2u & \Rightarrow x = \boxed{20} \end{cases} \end{aligned}$

The problem can be solved by analyzing the expression in the form $a^2$ + 2ab + $b^2$ .

Here it can be seen, $(2^4)^2$ + $2$ . $2^4$ . $2^{10}$ + $2^x$

Since here b= $2^{10}$ Therefore $b^2$ should be $2^{20}$ .

Thus x=20.

For it to be a square, 15 = 1+(8/2)+(x/2) and hence x = 20. This is due to the expansion of (2^4+2^x/2)^2