The answer to the second part of the question (in bold) is, `FindRoot` cannot provide answers to `Q2N[x] == 26`, because `Q2N` (and, equivalently, `Q2N1`) is not a differentiable function.  However, a solution clearly exists, which is the whole point of Cantor Pairing.  In fact, `Solve` provides the answer.

    Solve[26 == (n + d + 1) (n + d)/2 + d && n > 0 && d > 0, {n, d}, Integers]
    (* {{n -> 1, d -> 5}} *)

where `n` and `d` are shorthand for `Numerator[x]` and `Denominator[x]`.  The result is as expected.

More difficult cases are

    Q2N[.21]
    (* 7481 *)
    Solve[7481 == (n + d + 1) (n + d)/2 + d && n > 0 && d > 0, {n, d}, Integers]
    (* {{n -> 21, d -> 100}} *)

    Q2N[.213]
    (* 737291 *)
    Solve[737291 == (n + d + 1) (n + d)/2 + d && n > 0 && d > 0, {n, d}, Integers]
    (* {{n -> 213, d -> 1000}} *)

    Q2N[.2112]
    (* 287528 *)
    Solve[287528 == (n + d + 1) (n + d)/2 + d && n > 0 && d > 0, {n, d}, Integers]
    (* {{n -> 132, d -> 625}} *)

again as expected.  (`N[132/625]` is, indeed, precisely `0.2112`.)

**Plot of First 1000 Q2N**

For completeness, the solutions for the first 1000 values of `Q2N` can be plotted.

    ListPlot[Table[{i, n/d /. Flatten@Solve[i == (n + d + 1) (n + d)/2 + d && n > 0 && d > 0,
       {n, d}, Integers]}, {i, 1000}], AxesLabel -> {"Q2N[x]", x}]

[![enter image description here][1]][1]

Note that a solution does not exist for every `i`, although one does for most `i`.


  [1]: https://i.sstatic.net/x7V10.png