Since the feasible discrete set `RealAbs[Sqrt[2] - x / y] < 0.0001` is infinite, this is a hard task for `NMaximize`. It can be done as follows. First, we find the maximum for `x >= 1000 && y >= 1000` by

    NMaximize[{Log[1/y, RealAbs[Sqrt[2] - x/y]], 
    Element[{x, y}, Integers] && x >= 1000 && y >= 1000 && 
    RealAbs[Sqrt[2] - x/y] < 0.0001}, {x, y},Method->{"DifferentialEvolution", "ScalingFactor" ->3}]
which produces `{1.51158, {x -> 1649, y -> 1166}}` and a warning

> NMaximize::incst: NMaximize was unable to generate any initial points satisfying the inequality constraints {-0.0001+RealAbs[Sqrt[2]-Round[x]/Round[<<1>>]]<=0}. The initial region specified may not contain any feasible points. Changing the initial region or specifying explicit initial points may provide a better solution.

The above result is a feasible solution in view of

    N[RealAbs[Sqrt[2] - x/y] /. { x -> 1649, y -> 1166}]

> `0.0000231443`

Second, in order to be sure, we apply the known lower estimate for `RealAbs[Sqrt[2] - x/y]` (see the "Liouville numbers and transcendence" section in [that Wiki article](https://en.wikipedia.org/wiki/Liouville_number#Liouville_numbers_and_transcendence) for info) $$\exists A>0\, \forall\{x,y\}\in\mathbb{N} \left| \sqrt{2}-\frac{x}{y}\right|> \frac A {y^2} $$. According to Lemma from this section, we may take $A=\frac 1 {5}$ ($M=5$). This inequality implies the inequality  `ForAll[{x,y},{x,y} \[Element] PositiveIntegers && y>1, Log[1/y, RealAbs[Sqrt[2] - x/y]] < Log[1/y, 1/5] + 2]`. It is clear the maximum of ` Log[1/y, 1/5] + 2]` for all integers `y >=1000 ` is attained at `y==1000` and is equal to `N[Log[1/1000, 1/5]]==0.23299`.

Third, now we apply counting for `x>=1&&x<=999&&y>=2&&y<=999`

    Do[If[RealAbs[Sqrt[2] - x/y] < 0.0001, Print[{x, y, N[Log[1/y, RealAbs[Sqrt[2] - x/y]]]}]],
     {x, 1, 99}, {y, 2, 99}]

> `{99,70,2.24473} {140,99,2.07542}... {997, 705, 1.59229}`

Summarizing the above, we draw the conclusion that the maximum under consideration is equal to  `2.224473` at `{x==99,y==70}`.

Edit. `1000` in the above instead of `100` and constant $A= \frac 1 5$ instead of $A=\frac 1 3$.