In the usual setup, the probabilities of detection are related to the transmitivity and reflectivity of the beam splitter (ignoring phases for this discussion) so you cannot get 0 probability at one detector unless you remove the beam splitter or place a full mirror in your interferometer (reflectivity $0$ or $1$).
In general, if you detect photons one a the time, yes indeed you will detect all photons provided there is no loss and your detectors have 100% efficiency: photons will appear individually at one or the other detector. If you measure a classical beam, you can check via energy conservation (or more accurately power) that indeed there is no loss.
Here are some details for the single photon input case:
A 2-channel MZ device effects the unitary transformation
$$
U=\left(
\begin{array}{cc}
e^{-\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right) &
-e^{-\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta
}{2}\right) \\
e^{\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) &
e^{\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right)
\\
\end{array}
\right)
$$
between modes. Thus, acting on an input state with one photon in the first mode:
$$
a_1^\dagger\vert 0\rangle \equiv \vert 1\rangle \mapsto \left(\begin{array}{c} 1\\0\end{array}\right)
$$
we get the superposition
$$
\left(
\begin{array}{c}
e^{-\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right)
\\
e^{\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) \\
\end{array}
\right) = e^{-\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right)\vert 1\rangle + e^{\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) \vert 2\rangle
$$
where $a_2^\dagger\vert 0\rangle\equiv \vert 2\rangle$ is the state with one photon in the second mode.
Thus, the probability of finding the input state in the 2nd mode is $\sin^2(\beta/2)$, which is $0$
only if the transmittivity $\cos^2(\beta/2)=1$. In this case, the interferometer is then equivalent to a relative phase shifter:
$$
U\vert_{\beta=0}=\left(
\begin{array}{cc}
e^{-\frac{1}{2} i (\alpha +\gamma )} & 0 \\
0 & e^{\frac{1}{2} i (\alpha +\gamma )} \\
\end{array}
\right)
$$
and there is no splitting.
A similar analysis shows that the probability of getting $0$ in the first mode corresponds to $\beta=\pi$, in which case the input state is completely reflected to mode $2$ but accumulates an extra $\hbox{e}^{\frac{1}{2} i (\alpha -\gamma )}$ phase.
One often-used configuration is one where a 50-50 beamslitter is followed by a phase shifter and followed by a "reverse" 50-50 beam splitter. In matrix form this is:
$$
\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)\cdot
\left(
\begin{array}{cc}
e^{-\frac{i \alpha }{2}} & 0 \\
0 & e^{\frac{i \alpha }{2}} \\
\end{array}
\right)\cdot \left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)=\left(
\begin{array}{cc}
\cos \left(\frac{\alpha }{2}\right) & -i \sin
\left(\frac{\alpha }{2}\right) \\
-i \sin \left(\frac{\alpha }{2}\right) & \cos
\left(\frac{\alpha }{2}\right) \\
\end{array}
\right)
$$
which is an adjustable beamsplitter. One can vary the phase $\alpha$ by varying the optical path length in one arm so that $\cos(\alpha/2)=1$ (nothing reflected) or $\cos(\alpha/2)=0$ (nothing transmitted).
(A sketch by the OP would be useful to clarify if this is what they have in mind.)
Either way the previous argument runs the same: for detectors with 100% efficiency, the individual photons entering via input channel $1$ will be detected in one or the other detector, depending on $\alpha$.
