Looking at the schematic, and then at the datasheet (thanks Hyperion ! ), these type of diodes serve to set fixed voltage drop levels, and as per datasheet around 1.5~1.7 Volt (as a function of the current drawn, which can result into below 1.5V and above 1.7V).
This is confirmed in the schematic, when looking at D505:
+51.8V on the positive side (left of R519)
+50.0V on the negative side
i.e. dV = 1.8V
For D507:
-48.9V (-49.4V + 0.5V) on the positive side
-50.7V on the negative side
i.e. dV = 1.8V
LED technology is used to create a single PN junction providing the set voltage drop.
This is easiest replaced by 3x 1N4148 in series
Hence, using a 3.9V zener (which are noisy by the way) is more than doubling the required voltage drop.
D507 is in fact "enforcing" the current through R581/583, by imposing a fixed voltage drop across them.
Taking a fixed voltage drop of 0.5V across the B-E junction of the transistor, the voltage across the resistors is forced to 1.3V, and hence the current to ~4mA (I=U/R)
As a consequence, the current through the transistor is forced to 4mA.
By using a 3.9V zener, the current through the transistors is increased proportionally !
And thus their heat dissipation (P=dVxI)
It is not the right replacement.....
So, I would replace the diode by 3x 1N4148 in series