29. There exists a positive integer $m$. For example, when $m=12$, the indeterminate equation
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y z}=\frac{12}{x+y+z}$$

has infinitely many positive integer solutions.
After simplifying and rearranging equation (8), we get
$$x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)+x+y+z-9 x y z=0 .$$

Notice that, $(x, y, z)=(1,1,1)$ is a solution to (9). Suppose $(a, y, z)$ is a solution to (9) satisfying $a \leqslant y \leqslant z$, and consider (9) as a quadratic equation in $x$. Then $(b, y, z)=\left(\frac{y z+1}{x}, y, z\right)$ is also a positive real solution to (9), where $\frac{y z+1}{x}$ is obtained by Vieta's formulas, and $\frac{y z+1}{x} \geqslant z+\frac{1}{x}>z$. Since (9) is symmetric in $x, y, z$, $\left(y, z, \frac{y z+1}{x}\right)$ is a positive real solution to (9) satisfying $y \leqslant z<\frac{y z+1}{x}$. If we can prove that every term of the sequence $\left\{a_{n}\right\}$ defined below is a positive integer, then using the above derivation, we can find infinitely many positive integer solutions to (9). Thus, when $m=12$, (8) has infinitely many positive integer solutions.

The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{0}=a_{1}=a_{2}=1, a_{n+2}=\frac{a_{n} a_{n+1}+1}{a_{n-1}}, n=1,2, \cdots$.
To prove that every term of $\left\{a_{n}\right\}$ is a positive integer, we can use mathematical induction to prove that the following conclusions hold simultaneously:
(1) For any $n \in \mathbf{N}^{*}$, $a_{n} \in \mathbf{N}^{*}$;
(2) $a_{n-1}\left|\left(a_{n} a_{n+1}+1\right), a_{n+1}\right|\left(a_{n-1} a_{n}+1\right)$;
(3) $a_{n} \mid\left(a_{n+1}+a_{n-1}\right)$.

The specific derivation process is left to the reader.