Solution:
Preliminary remark: In what follows let $n$ be odd. Without loss of generality $a>b, c>d, e>f, g>h$. The condition equation is then equivalent to
$$n 2^{d+h-b-f}\left(2^{c-d}-1\right)\left(2^{g-h}-1\right)=\left(2^{a-b}-1\right)\left(2^{e-f}-1\right).$$
Since $n$ is odd, we have $d+h-b-f=0$, and one can, without loss of generality, set $d=h=b=f=0$:
$$n\left(2^{c}-1\right)\left(2^{g}-1\right)=\left(2^{a}-1\right)\left(2^{e}-1\right)$$

1st Solution (sketched): The numbers of the form $n=19+64k$ with integer $k \geqq 0$ cannot be represented. For $x>1$, $2^{x}-1$ gives remainder 3 upon division by 4, for $x=1$ remainder 1. Only if an odd number of the variables $a, c, e, g$ has value 1 can equation (*) be satisfied. If three of the variables have value 1, the equation cannot hold, since $n$ does not have the form $2^{x}-1$. Hence exactly one of the brackets in $(*)$ has value 1. By corresponding considerations and case distinctions concerning remainders upon division by 8, 16, 32, 64, one can fix the values of the remaining brackets and in each case lead to a contradiction.

2nd Solution (sketched): The number $n=19$ is not representable. One uses the relation $\gcd(2^{x}-1, 2^{y}-1)=2^{\gcd(x, y)}-1$ to show: from $(2^{x}-1) \mid (2^{y}-1)(2^{z}-1)$ it follows that $x \mid y$ or $x \mid z$.
From this it follows, together with $(*)$, that 19 can be written as
$$19=\frac{2^{a}-1}{2^{c}-1} \cdot \frac{2^{e}-1}{2^{g}-1} \quad \text{or} \quad 19=\frac{2^{a}-1}{(2^{c}-1)(2^{g}-1)} \cdot (2^{e}-1)$$
as a product of two integer factors. Since 19 is a prime number and does not have the form $2^{e}-1$, it suffices to show that the equation
$$19=\frac{2^{a}-1}{(2^{c}-1)(2^{g}-1)}$$
cannot hold. The numerator contains the prime factor 19, which is only possible for $a \geq 18$. The numbers $c$ and $g$ divide $a$, without loss of generality $c \geq g$. For $c=g=a/2$ the fraction has the non-integer value $1+2/(2^{a/2}-1)$, for $c \leq a/3, g \leq a/3$ or for $c=a/2, g \leq a/4$ the value is too large. Thus $c=a/2, g=a/3$, and the fraction has the non-integer value $2^{a/6}+1/(2^{a/6}-1)$.

Remark: It is not always the case that $n$ is a product of two integer factors of the form $(2^{x}-1)/(2^{y}-1)$. For example, $13=\frac{(2^{12}-1)(2^{2}-1)}{(2^{6}-1)(2^{4}-1)}$, but one can show, as in the 2nd Solution, that 13 cannot be represented in the form $(2^{x}-1)/(2^{y}-1)$.