Since QM is probabilistic, you can only know probability until a measurement is performed.
Your doubt can be referred to a "1-input" beam-splitter since you're dealing only the branches, without analyzing the whole interferometer.
Now, if you trust this probabilistic interpretation, we can say avoiding any math,
- that a photon may pass through all the branches (with probability determined by the beam-splitter itself, that can be represented with a reflection, $r$ and a transmission coefficient, $t$, not independent);
- that a photon has passed through a determined branch after you have measured it.
Math model. Let's recap this interpretation without dealing with Mach-Zehnder interferometer but just focusing on a beam-splitter without any detail about the EM field and phase, since your doubt is about the presence of the photon in one or the other branch.
Being the branch taken by the output photon a measurable physical quantity, it can be represented by an Hermitian operator $\hat{B}$ whose (orthonormal) eigenstates $\{|\psi_i\rangle\}_{i=1,2}$ represent the state of the system if the photon takes the $i^{th}$ branch, while the eigenvalues $b_i$ are the labels of the branches. The state of the photon after the 1-input beam-splitter can be thus written as a superposition of these eigenstates,
$$|\psi\rangle = c_1 |\psi\rangle_1 + c_2 |\psi\rangle_2 \ ,$$
being the coefficients $c_i$ determined by the very nature of the beam splitter (e.g. 1: transmission $c_1 = t$, 2: reflection, $c_2 = r$) and determining the probability for the photon taking the $i^{th}$ branch, $p_i= c_i^* c_i = |c_i|^2$. Normalization condition reads
$$1 = \langle \psi | \psi \rangle = \langle c_1 \psi_1 + c_2 \psi_2 | c_1 \psi_1 + c_2 \psi_2 \rangle = |c_1|^2 + |c_2|^2 = t^2 + r^2 \ ,$$
showing that transmission and reflection are not independent.
As an example, numbering branches with natural numbers $\{b_1, b_2\} = \{0,1\}$ the expected value of the branch is evaluated as
$$\langle \hat{B} \rangle = \langle \psi | \hat{B} | \psi \rangle = \dots = |c_1|^2 b_1 + |c_2|^2 b_2 = r^2 \ ,$$
with the choice done.
Remarks. For a non-ideal beam-splitter some photons may be absorbed, so absorption need to be considered as well in the transition from input to output.
