If methanol and nitromethane form an azeotrope, then you'll never obtain the pure nitromethane. An azeotrope behaves as a single compound, which boils at a certain temperature, which is above both boiling points of the pure compounds, which constitute the azeotrope.
An example, which I know quite well. HNO3 and H2O form an azeotrope with 68% HNO3 and 32% H2O by weight. HNO3 boils at around 80 C. H2O boils at 100 C. If you mix H2O with HNO3 in the azeotropic ratio, then that liquid boils at around 120 C. So, heating up to 80 C or 100 C does not give any boiling and also does not give an appreciable rate, at which a vapor is produced.
If you have a mix with excess water, then that mix will first start boiling at a temperature between 100 and 120 C, giving off mainly water. While the water content decreases, the boiling point will increase and also the relative amount of HNO3 in the water/HNO3 mix will increase. This continues, until the concentration of water is so low, that the azeotropic ratio is reached. At that point the maximum boiling point is reached and the vapor also contains H2O and HNO3 in the azeotropic ratio. A similar story can be told for excess HNO3 in the mix. Then first relatively much HNO3 is boiled over at a temp lower than 120 C. While the concentration of water increases, the boiling point increases.