First, I would reword the question to "Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?"
Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:
A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.
A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.
So, the spectrum is not sensitive, the fact is that temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.