Example 1: Does there exist a geometric sequence of positive integers with only 5 terms whose sum is 211? If it exists, determine this sequence; if not, please explain the reason.
If there exists a geometric sequence that satisfies the condition,  
let it be $\frac{a}{q^{2}}+\frac{a}{q}+a+a q+a q^{2}=211$.  
$\because$ each term of the sequence is a positive integer,  
$\therefore q$ is a rational number, $q=\frac{c}{d}(c, d$ are coprime), the equation becomes:  
$$
\frac{a d^{2}}{c^{2}}+\frac{a d}{c}+a+\frac{a c}{d}+\frac{a c^{2}}{d^{2}}=211 \text {. }
$$
$\because$ each term is a positive integer, $\therefore c^{2}\left|a, d^{2}\right| a, a=$ $k c^{2} d^{2}, k \in \mathbf{N}^{*}$, thus $k$ divides the left side of (2), giving: $k \mid 211$.  
$\because 211$ is a prime number,  
$\therefore k=1, a=c^{2} d^{2}$, at this point (2) becomes:  
$d^{4}+d^{3} c+d^{2} c^{2}+d c^{3}+c^{4}=211$  
(3) is an indeterminate equation.  
$$
\because 4^{4}=256>211, \therefore c, d<4 .
$$

If one of $c, d$ is 1, such as $c=1$, then $d^{4}+d^{3}+d^{2}+d=210$, but $d \neq 2,3 . \therefore c, d$ must be one 2 and the other 3, at this time $3^{4}+3^{3} \cdot 2+3^{2} \cdot 2^{2}+$ $3 \cdot 2^{3}+2^{4}=211, a=36, q=\frac{2}{3}$ or $\frac{3}{2}$.