Why can the neutron not be measured in this fixed-target experiment?

Exclusive & semi-inclusive reactions are commonly as written target(beam, detected)undetected:

$$p(pi^-, pi^+)npi^-$$

To understand the detection of the final state, we really need more information. While there are detectors that can handle multiple particle final states (https://www.jlab.org/physics/hall-b/clas), this experiment wasn't done there.

More commonly, fixed target experiments put a beam on target and look at reaction products with a spectrometer (e.g., SLAC's 8 GeV spectrometer: https://www.slac.stanford.edu/pubs/slacpubs/5750/slac-pub-5753.pdf). This grabs a small bite of both angular space and momentum space. It uses (a) dipole magnet(s) to select a central momentum, and quadruple magnets to focus particle tracks so that momentum and lab-angle (regardless of out-of-plane angle) can be determined.

The spectrometer can only look a one charged final state, but that can be reversed by reversing the current. Since the beam is $pi^-$, the $pi^+$ final state is much cleaner.

The way to separate the reaction from:

$$p(pi^-, pi^+)X$$

is the invariant mass of $X$. The initial state's 4-momentum is:

$$p_i^{mu} = p_p^{mu} + p_p^{pi^-} $$ $$p_i^{mu} = (M_p, 0,0,0) + (E_{pi^-}, 0,0,p_{pi^-}) $$

while the final state is:

$$p_f^{mu} = p^{mu}_{pi^+} + p_X^{mu} $$

Of course, $p_i^{mu}= p_f^{mu}$ so that:

$$p_X^{mu} = p_i^{mu} - p^{mu}_{pi^+} $$

$$p_X^{mu} = (M_p+E_{pi^-}, 0,0, p_{pi^-}) - (E_{pi^+}, p_{pi^+}sin{theta},0, p_{pi^+}cos{theta}) $$

If you work that all out at a fixed angle, there should be a momentum region that ensures $X$ is a nucleon plus a pion, and since it has charge -1, it has to be a neutron and $pi^-$. (Another reason to detect the positive pion, as a negative pion in the final state does NOT distinguish between an undetected $npi^+$ and $ppi^-$). That also rules out a lot of other final states, so by detecting the $pi^+$, there is a kinematic region where $X=npi^-$ is the only viable option.