How to justify this contradiction in acid-base equilibrium of methanol and water?
We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding. [...]

[...] It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).



You are pointing out competing factors that, in abstract, should lead to either methanol or water being more acidic than the other (that is. lead to either hydroxide or methoxide to be more basic than the other, respectively). Once we move from considering properties in abstract to an actual system in which an acid-base equilibrium is present, the competing factors are superimposed, with a non-contradictory result: either methanol will be more acidic (and hydroxide more basic) or water will be more acidic (and methoxide more basic), depending on which factors dominate. (In particular, as pointed out in Why is methanol more acidic than water? and Is methanol really more acidic than water?, the relative acidities depend on the solvent.)