Why is phenol more acidic than aniline? I find that there are two contradicting factors at play here:

1) On removing proton, negative charge on oxygen is more stable than it is on nitrogen (because of greater electronegativity of oxygen). Therefore, phenol is more acidic than aniline.

2) +M effect of nitrogen is more than oxygen, so negative charge (after deprotonation) in the case of aniline will be more delocalized making the anion more stable. Therefore, aniline should be more acidic than phenol.

So is that a factual question or am I missing something?
Taken from my answer here:

Comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units
Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm p K_\mathrm a$ difference. Resonance in triphenyl system is effectively only one phenyl group at a time due to the steric effects due to the bulkiness of the phenyl groups.

Here methane and acetylene form anions where the negative charge on an $\mathrm{sp^3}$ and $\mathrm{sp}$ carbon respectively. Here, there is an electro negativity difference between the two carbons which leads to a very high difference in $\mathrm p K_\mathrm{a}$.

The difference between their electronegativities too large to explain the $\mathrm p K_\mathrm a$ difference of $15$. Nitrogen is $3$ and oxygen is $3.5$ on Pauling scale of electronegativity.

This small amount makes a big difference in $\mathrm p K_\mathrm a$ values. I say this because, if methane and acetylene have a difference of 20 units where the only difference is their hybridization and therefore electronegativity, oxygen and nitrogen will have a big difference in $\mathrm p K_\mathrm{a}$ values.
This can be seen in the actual values as referenced in Evan's $\mathrm pK_\mathrm a$ table, where aniline has a $\mathrm pK_\mathrm a$ value of $30.6$ and phenol has a $\mathrm p K_\mathrm a$ value of $18.0$ both referenced in DMSO.