It's Not Infinity

$\begin{aligned} y & = \log_2 x + \log_4 x + \log_{16} x + ... \\ & = \log_2 x + \frac {\log_2 x}{\log_2 4} +\frac {\log_2 x}{\log_2 16} + ... \\ & = \log_2 x + \frac 12 \log_2 x + \frac 14 \log_2 x + ... \\ & = \log_2 x \left(1 + \frac 12 + \frac 1{2^2} + \frac 1{2^3} + ... \right) \\ & = \log_2 x \left(\frac 1{1-\frac 12} \right) \\ & = 2 \log_2 x \end{aligned}$

Now, we have:

$\begin{aligned} 4\log_4 x & = \frac {\sum_{k=1}^y (4k+1)}{\sum_{k=1}^y (2k-1)} \\ 4 \cdot \frac {\log_2 x}{\log_2 4} & = \frac {4\cdot \frac {y(y+1)}2 +y}{2\cdot \frac {y(y+1)}2 -y} \\ 2 \log_2 x & = \frac {2y^2+3y}{y^2} \\ y & = \frac {2y+3}y \\ y^2 & = 2y+3 \end{aligned}$

$\begin{aligned} \implies y^2 - 2y -3 & = 0 \\ (y+1)(y-3) & = 0 \\ \implies y & = 3 & \small \color{#3D99F6}{y = -1 < 0 \text{ is not acceptable.}} \end{aligned}$

$\begin{aligned} \implies 2 \log_2 x & = y = 3 \\ x & = 2^\frac 32 \\ \implies x^2y & = \left(2^\frac 32\right)^2 \cdot 3 = 2^3 \cdot 3 = \boxed{24} \end{aligned}$