Do interferometers require equal-powered (or as equal as possible) beams being emitted at the target and the photodetector(s)?

It's actually a little bit of a tricky design question.

The rough story is this. In principle, it doesn't matter at all. You will see interference fringes even if the beams have different powers. In practice, it can matter, depending on how small of a signal you are trying to measure.

It will depend on the details of your photodetector. There are a few signal levels that matter. Your photodetector outputs a voltage $V$ proportional to the instantaneous power $P$ on the photodetector. There is a $V_{text{noise}}$ corresponding to the electronic noise floor of the photodetector. There is a $V_{text{max}}$ corresponding to the max voltage the photodetector can output before it saturates.

If you have two beams that result in voltages $V_1$ and $V_2$ on the photodetector then when you put both beams on together you will get a signal like

$$ V = V_1 + V_2 + 2sqrt{V_1V_2}cos(phi) $$

Where $phi$ is the phase between the two beams of light that varies as a function of whatever you are measuring about your "target"

In addition to electronic noise you will also get optical shot noise which is proportional to $sqrt{V}$.

So to resolve your interference fringes you want your signal, proportional to $sqrt{V_1V_2}$, to be as large as possible. BUT, you are limited by $V<V_{text{max}}$ so that you don't saturate the detector.

The contrast of an interference fringe given by $S = A + Bcos(phi)$ with $0<B<A$ is given as

$$ C = frac{2B}{A+B} $$

$Crightarrow 0$ for small signal if $Bll A$ and $Crightarrow 1$ for "perfect contrast" with $B=A$.

In this case we want to maximize $B$ with the constraint that $A+B$ is fixed at $V_{text{max}}$. This occurs when our signal has maximum contrast such that $B=A$. In the case of your problem this means that you, ideally, would have $V_1=V_2$.

That is, you want the power from beam 1 to equal the power from beam 2 on the photodetector. This may mean that you need more power in the arm including the target if the target scatters out or absorbs any of the light which falls on it.

All of this said, if you are measuring something "easy" then it is likely you will not need to worry about all of these fine details of signal to noise and you will be fine with imbalanced powers. Tricky say without doing a full design analysis. Everything I describe here will probably get you a factor of 2 or 4 or something in signal-to-noise which may or may not make the difference for your application.

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edit: The OP asks about a case when the power which can go to the sample is limited. For example the sample might be damaged in some way if it has too much power on it. In a single photodetector interference setup this is not ideal. Absent electronic noise on the photodetector (or anywhere in the detection chain including amplifiers or data acquisition devices) the best strategy would be to set the two power on the detector equal as described above. This basically optimizes amplification of the small signal with shot noise.

However, in the presence of electronic noise it is likely that the two low power levels on the photodetector will not be enough to exceed electronic noise. In this case you need boost the power in the "local oscillator" (LO) beam to amplify the sinusoidal modulation above the electronic noise floor. You should increase the LO power until you reach saturation of the photodetector. But I'll add the caveat that once shot noise from LO exceeds the electronic noise floor you don't see any improvement in signal to noise for increases in LO power. Basically both shot noise and signal level increase $propto sqrt{V_{text{LO}}}$ in that case.

However, the BEST strategy, if you must perform an interferometrically enhanced measurement of a low level signal, is to use balanced interferometric detection. In this case you use two photodetectors to measure the light out of each port of the beamsplitter and then you subtract the two signal before amplification. In this case your signal appears as a modulation about zero volts, i.e. there is no DC offset. This allows you to use a much larger LO power without saturating the photodetectors making it easy to boost the small signal beyond the electronic noise floor of the detector chain. That said, once the signal is boosted beyond the electronic noise floor, you again don't get further gains in SNR as you increase the LO more.