Positive reals $x, y, z$ satisfy $$\frac{xy+1}{x+1} = \frac{yz+1}{y+1} = \frac{zx+1}{z+1}$$

Do they all have to be equal?

[i](Proposed by Oleksii Masalitin)[/i]
Given the equations:
\[ \frac{xy+1}{x+1} = \frac{yz+1}{y+1} = \frac{zx+1}{z+1} \]

Let us denote the common value by \( k \). Therefore, we have:
\[ \frac{xy+1}{x+1} = k \]
\[ \frac{yz+1}{y+1} = k \]
\[ \frac{zx+1}{z+1} = k \]

From the first equation:
\[ xy + 1 = k(x + 1) \]
\[ xy + 1 = kx + k \]
\[ xy - kx = k - 1 \]
\[ x(y - k) = k - 1 \]
\[ x = \frac{k - 1}{y - k} \]

From the second equation:
\[ yz + 1 = k(y + 1) \]
\[ yz + 1 = ky + k \]
\[ yz - ky = k - 1 \]
\[ y(z - k) = k - 1 \]
\[ y = \frac{k - 1}{z - k} \]

From the third equation:
\[ zx + 1 = k(z + 1) \]
\[ zx + 1 = kz + k \]
\[ zx - kz = k - 1 \]
\[ z(x - k) = k - 1 \]
\[ z = \frac{k - 1}{x - k} \]

Now, we have:
\[ x = \frac{k - 1}{y - k} \]
\[ y = \frac{k - 1}{z - k} \]
\[ z = \frac{k - 1}{x - k} \]

Substituting \( y \) and \( z \) into the expression for \( x \):
\[ x = \frac{k - 1}{\frac{k - 1}{z - k} - k} \]
\[ x = \frac{k - 1}{\frac{k - 1 - k(z - k)}{z - k}} \]
\[ x = \frac{k - 1}{\frac{k - 1 - kz + k^2}{z - k}} \]
\[ x = \frac{k - 1}{\frac{k^2 - kz}{z - k}} \]
\[ x = \frac{k - 1}{k^2 - kz} \cdot (z - k) \]
\[ x = \frac{(k - 1)(z - k)}{k^2 - kz} \]

Since \( x = \frac{k - 1}{y - k} \), we can equate:
\[ \frac{k - 1}{y - k} = \frac{(k - 1)(z - k)}{k^2 - kz} \]

This implies:
\[ y - k = \frac{k^2 - kz}{z - k} \]
\[ y - k = k \]

Thus, \( y = k \). Similarly, we can show that \( x = k \) and \( z = k \).

Therefore, \( x = y = z \).

The final answer is \( \boxed{ x = y = z } \).