Find min and max of magnitude ltspice
The voltage you have there is the result of a .TRAN simulation, and its values alternate from negative to positive all across the timescale. Trying to find a maximum or a minimum will only get you the ± peak values (i.e. about ±6 V from your graph), which is not what you want. BTW, mag() is used in .AC, only.
I see a waveform in blue. If that waveform is known (not just plotted for comparison), then you can use that for measuring, or if you have a demodulator then you can use the filtered waveform, but you should account for possible phase delays and attenuations. This means it's not perfect, but easily done:
.meas vmax max v(x)
.meas vmin min v(x)

Or, if you know the frequency of the modulating signal and the carrier (and their phases), then you can use those informations to measure the approximate max and min. I say &quot;approximate&quot; because for e.g. fm=1 Hz, fc=10 Hz you know that fc is an even number and, thus, its peaks will not coincide with the peaks from fm, unless fc is a cosine. But for fc=10, the peaks that can be detected are (considering zero phase for both):
.meas div param 1/4/fc  ; nr of time divisions is 4x the carrier (4 quadrants)
.meas t1 param (4*2-1)/div  ; 4 half periods - 1 quadrant
.meas t2 param (12*2-1)/div ; 12 half perionds - 1 quadrant
.meas vmax find abs(v(x)) at t1
.meas vmin find abs(v(x)) at t2

Otherwise it's next to impossible, particularly if not even the modulating frequency is known (but which could be measured), since it would imply measuring peak by peak, several measurements (e.g. think 433 kHz and 1 kHz) which means several (many) .meas lines, then doing the finite difference of these lines (another maximum n-1 lines), to find where the zero is, and this only for the maximum -- thus twice for the minimum.
For these reasons, if measurements are needed, it's better to come prepared with predefined values so that the measurements can be done easily (the first two cases). If the modulator works as intended, then you know the measurements will hold for any other frequency. So, with a known fm, add a known fc and then measure for testing.
All the above rely on .meas scripts, but that's not the only way of doing it. You can also use four. For the numbers above, this will give you the sidebands and the carrier:
.four {fm} {(fm+fc)/fm} {N} v(x)  ; N = nr of periods

You could add your own circuit that performs AM detection, but that would mean something similar to the first part; no need to encumber the solver with twice the same job.