We can use the expression that is commonly used to estimate the apparent magnitude of a planet or asteroid in the Solar System:
$$\boxed{m=5 \log \frac{1329}{d \cdot \sqrt p}+5 \log (D_s \cdot D)-2.5\log f(F)}$$
Where:
m is the apparent magnitude
log is the decimal logarithm
d is the diameter in km
p is the albedo
Ds is the distance to the Sun in Astronomical Units (AU)
D is the distance Earth-Planet in Astronomical Units (AU)
F is the phase angle
f(F) is the phase function
As a phase function we will use the usual one:
$$f(F)=\dfrac{1+\cos F}2$$
1. Magnitude of Mars. Opposition (maximum)
$d=2\cdot 3389.5$ km
$p=0.15$
$D_s = 1.381497$ AU. Mars Perihelion
$D=1.381497 - 1.016714$ AU. Mars Perihelion and Earth Aphelion
$F=0$ Opposition
$$m \simeq -2.97$$
2. Magnitude of the Earth in quadrature = greatest elongation as seen from Mars
$d=2\cdot 6371$ km
$p=0.367$
$D_s = 1.016714$ AU. Earth Aphelion
$D=\sqrt{1.381497^2 - 1.016714^2 \ }$ AU. Mars Perihelion and Earth Aphelion
$F=\dfrac{\pi}2$ Earth at greatest elongation as seen from Mars
$$m \simeq -3.21$$
Probably the greatest brightness of the Earth as seen from Mars will not be when the Earth is exactly in quadrature, just as the greatest brightness of Venus as seen from the Earth is not exactly at quadrature but when it is somewhat closer to the Earth.
But if already with the Earth in quadrature we obtain $m \simeq -3.21$ that is a greater brightness than $m \simeq -2.97$ we can claim (according to this estimation), that the maximum brightness of the Earth as seen from Mars is slightly higher than the maximum brightness of Mars as seen from the Earth, although the difference between the both is not large.
A similar thread: Formula to calculate the apparent magnitude of Earth from arbitrary distances
Best regards.