Given

$\rm \frac{2^{x+1}}{16} + \frac{16}{2^{x + 1}} = \frac{65}{8}$

$\rm or, \frac{2^{x+1}}{16} + \left ( \frac{2^{x+1}}{16} \right )^{-1} = \frac{65}{8}$ 

Let $\rm a = \frac{2^{x + 1}}{16}$, we get,

$\rm or, a + a^{-1} = \frac{65}{8}$

\[\rm a + \frac{1}{a} = \frac{65}{8}\]

Multiplying both sides by \(\rm a\), we get:

\[\rm a^2 + 1 = \frac{65a}{8}\]

Multiplying both sides by \(\rm 8\), we get:

\[\rm 8a^2 + 8 = 65a\]

Rearranging terms, we have a quadratic equation in \(\rm a\):

\[\rm 8a^2 - 65a + 8 = 0\]

The above equation is quadratic in a. We use the mid-term factorization formula to solve the equation.

$\rm or, 8a^{2} - (64 + 1) a + 8 = 0$

$\rm or, 8a^{2} - 64a - a + 8 = 0$

$\rm or, 8a (a - 8) - 1 (a - 8) = 0$

$\rm or, (8a - 1)(a - 8) = 0$

Either

$\rm (8a - 1) = 0$

$\rm or, 8a = 1$

$\rm or, a = \frac{1}{8}$

Substitute $\rm a = \frac{2^{x + 1}}{16}$ into the equation, we get,

$\rm or, \frac{2^{x + 1}}{16} = \frac{1}{8}$

$\rm or, 2^{x + 1} = 2$

$\rm or, 2^{x + 1} = 2^{1}$

$\rm or, x + 1 = 1$

$\rm or, x = 1 - 1$

$\rm \therefore x = 0$

Or

$\rm (a - 8) = 0$

$\rm or, a = 8$

Substitute $\rm a = \frac{2^{x + 1}}{16}$ into the equation, we get,

$\rm or, \frac{2^{x + 1}}{16} = 8$

$\rm or, 2^{x + 1} = 16 \cdot 8$

$\rm or, 2^{x + 1} = 2^{4} \cdot 2^{3}$

$\rm or, 2^{x + 1} = 2^{4 + 3}$

$\rm or, 2^{x + 1} = 2^{7}$

$\rm or, x + 1 = 7$

$\rm or, x = 7 -1 $

$\rm \therefore x = 6$

Hence, the required values of x are x = {0, 6}.