The system of equations
\[\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 1\]
has exactly one solution. What is $z$ in this solution?
We can solve for $x,$ $y,$ $z$ separately.
First, $\frac{xy}{x + y} = 1$ simplifies to $xy = x + y.$ Adding 1 to both sides gives $xy + 1 = x + y + 1,$ which factors as $(x + 1)(y + 1) = 2.$ Since 2 is not the product of two integers that differ by 1, we have no solution for $x$ and $y.$
Next, $\frac{xz}{x + z} = 2$ simplifies to $xz = 2x + 2z.$ Then $xz - 2x - 2z = 0,$ which factors as $(x - 2)(z - 2) = 4.$ Thus, the solutions for $x$ and $z$ are the pairs
\[(x,z) = (2,6), \quad (x,z) = (3,-2), \quad (x,z) = (1,1).\]Since not all of these satisfy the third equation, $\frac{yz}{y + z} = 1,$ the unique system solution is $(x,y,z) = (1,1,3),$ so $z = \boxed{3}.$
To find the value of $z$ in the given system, we can start by multiplying each equation by the denominators to clear the fractions. This gives us:
\begin{align*}
xy &= (x + y), \\
xz &= 2(x + z), \\
yz &= (y + z).
\end{align*}Next, let's expand each equation to get:
\begin{align*}
xy &= x + y , \\
xz &= 2x + 2z , \\
yz &= y + z .
\end{align*}Rearranging terms, we can rewrite these equations as:
\begin{align*}
xy - x - y &= 0 , \\
xz - 2x - 2z &= 0 , \\
yz - y - z &= 0 .
\end{align*}Factoring the expressions on the left side of each equation, we have:
\begin{align*}
(x - 1)(y - 1) &= 1 , \\
(x - 2)(z - 2) &= 4 , \\
(y - 1)(z - 1) &= 1 .
\end{align*}Since the given equations have exactly one solution, it means that the expressions on the left side of each equation must all be equal. This gives us the following system of equations to solve:
\begin{align*}
(x - 1)(y - 1) &= (x - 2)(z - 2) , \\
(x - 1)(y - 1) &= (y - 1)(z - 1) .
\end{align*}From the first equation, we see that $(x - 1)(y - 1) = (x - 2)(z - 2)$. Dividing both sides by $(x - 1)$, we have:
\[y - 1 = \frac{x - 2}{x - 1} \cdot (z - 2) .\]Similarly, dividing both sides of the second equation by $(y - 1)$, we get:
\[1 = \frac{x - 1}{y - 1} \cdot (z - 1) .\]Multiplying the above two equations together, we have:
\begin{align*}
(y - 1)(1) &= \left( \frac{x - 2}{x - 1} \cdot (z - 2) \right) \left( \frac{x - 1}{y - 1} \cdot (z - 1) \right) \\
(y - 1) &= (z - 2)(x - 1) \\
y - 1 &= zx - 2z - x + 2 .
\end{align*}Rearranging the terms, we can write this equation as:
\[zx - x - y = 1 + 2z .\]Now, let's substitute the given values into this equation. From the first equation of the original system, we have $xy = x + y$, so $y = \frac{x}{x - 1}$. Substituting this into the equation above, we get:
\[zx - x - \frac{x}{x - 1} = 1 + 2z .\]Multiplying through by $(x - 1)$ to clear the fraction, we have:
\[(zx - x)(x - 1) - x = (x - 1)(1 + 2z) .\]Expanding both sides, we get:
\[x^2 z - 2x^2 - xz + x - x = x + 2xz - 2z - 1 - x + 2z .\]Simplifying, we have:
\[x^2 z - 2x^2 - xz = 2xz - 1 .\]Collecting like terms, we obtain:
\[x^2 z - 2x^2 - 3xz + 1 = 0 .\]To make further progress, let's write this equation in terms of $z$. Factoring the expression on the left side, we get:
\[x(xz - 2x - 3z) + 1 = 0 .\]Since the given system has exactly one solution, it means that $x$ is nonzero. Therefore, we can divide both sides by $x$ to find:
\[xz - 2x - 3z = -\frac{1}{x} .\]Rearranging terms, we can write this as:
\[xz - 3z = 2x - \frac{1}{x} .\]Factoring out $z$, we have:
\[z(x - 3) = 2x - \frac{1}{x} .\]Now, let's substitute the given values from the second equation of the original system. We have $xz = 2(x + z)$, so $x = \frac{2z}{z - 2}$. Substituting this into the equation above, we get:
\[z \left( \frac{2z}{z - 2} - 3 \right) = 2 \left( \frac{2z}{z - 2} \right) - \frac{1}{ \frac{2z}{z - 2} } .\]Simplifying, we have:
\[z \left( \frac{2z}{z - 2} - 3 \right) = \frac{4z}{z - 2} - \frac{z - 2}{2z} .\]Multiplying both sides by $(z - 2)$ and simplifying, we obtain:
\[2z^3 - 10z^2 + 12z - 4 = 0 .\]Factoring out a $2$ from all terms, we have:
\[2(z - 2)(z - 1)(z - 1) = 0 .\]Therefore, the equation $2z^3 - 10z^2 + 12z - 4 = 0$ factors as $2(z - 2)(z - 1)^2 = 0$. Setting each factor equal to zero, we find that $z - 2 = 0$ or $z - 1 = 0$. Therefore, $z = \boxed{2}$.