A triple $(x,y,z)$ of real numbers is called a [i]superparticular[/i] if
$$\frac{x+1}{x} \cdot \frac{y+1}{y}=\frac{z+1}{z}$$
Find all superparticular positive integer triples.
1. Start with the given equation for a superparticular triple:
   \[
   \frac{x+1}{x} \cdot \frac{y+1}{y} = \frac{z+1}{z}
   \]
   Simplify the left-hand side:
   \[
   \left(1 + \frac{1}{x}\right) \left(1 + \frac{1}{y}\right) = 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy}
   \]
   Thus, the equation becomes:
   \[
   1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} = 1 + \frac{1}{z}
   \]
   Subtract 1 from both sides:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} = \frac{1}{z}
   \]
   Multiply through by \(xyz\):
   \[
   yz + xz + xy = xyz
   \]
   Rearrange to get:
   \[
   z(x + y + 1) = xy
   \]

2. Let \(d = \gcd(x, y, z)\). Then we can write \(x = dx_1\), \(y = dy_1\), and \(z = dz_1\) where \(\gcd(x_1, y_1, z_1) = 1\). Substitute these into the equation:
   \[
   dz_1(d x_1 + d y_1 + 1) = d^2 x_1 y_1
   \]
   Simplify by dividing through by \(d\):
   \[
   z_1(d x_1 + d y_1 + 1) = d x_1 y_1
   \]
   Thus, \(d \mid z_1\). Let \(z_1 = d z_2\):
   \[
   d z_2(d x_1 + d y_1 + 1) = d x_1 y_1
   \]
   Simplify by dividing through by \(d\):
   \[
   z_2(d x_1 + d y_1 + 1) = x_1 y_1
   \]

3. Let \(\gcd(x_1, z_2) = a\) and \(\gcd(y_1, z_2) = b\). Since \(\gcd(x_1, y_1, z_2) = 1\), we have \(\gcd(a, b) = 1\). Thus, \(ab \mid z_2\). Replace \(x_1, y_1, z_2\) with \(a x_2, b y_2, ab z_3\) respectively:
   \[
   ab z_3(d a x_2 + d b y_2 + 1) = a x_2 b y_2
   \]
   Simplify by dividing through by \(ab\):
   \[
   z_3(d a x_2 + d b y_2 + 1) = x_2 y_2
   \]
   Since \(\gcd(z_3, x_2) = \gcd(z_3, y_2) = 1\), we get \(z_3 = 1\). Thus, we need to solve:
   \[
   d a x_2 + d b y_2 + 1 = x_2 y_2
   \]

4. Rearrange the equation:
   \[
   x_2 y_2 - d a x_2 - d b y_2 = 1
   \]
   Factorize:
   \[
   (x_2 - d b)(y_2 - d a) = d^2 a b + 1
   \]
   Let \(y_2 - d a = t\), then:
   \[
   x_2 = \frac{d^2 a b + d b t + 1}{t}
   \]

5. The solutions are:
   \[
   (x, y, z) = \left(d a \cdot \frac{d^2 a b + d b t + 1}{t}, d b (d a + t), d^2 a b\right)
   \]

The final answer is \( \boxed{ \left(d a \cdot \frac{d^2 a b + d b t + 1}{t}, d b (d a + t), d^2 a b\right) } \)