First, there is a mistake in the sign `k`. And secondly, for such tasks we use a special method that expands the possibilities of the shooting method. I will demonstrate the working code

    A1 = 200;
    A2 = 186*10^5;
    A3 = 1/5000;
    a = Rationalize[A2*A3^3];
    H = 1; r0 = 10^-5;
    f[z_, r_] := A1*(z + H) + a*(1/(Sqrt[z^2 + r^2] - 1))^3;
    k = -z''[r]/(z'[r]^2 + 1)^(3/2) - z'[r]/(r Sqrt[z'[r]^2 + 1]);
    sol = ParametricNDSolveValue[{k == f[z[r], r], z'[r0] == 0, 
        z[r0] == z0}, z, {r, r0, 1}, {z0}];
    Plot[Evaluate[Table[sol[z0][r], {z0, -1.15, -1.05, .01}]], {r, r0, 1},
      PlotRange -> All]

[![fig1][1]][1]

Find the parameter that satisfies the boundary condition on the right border

    FindRoot[sol[z0][1] == -H, {z0, -1.05}]

    Out[]= {z0 -> -1.03176}
    {Plot[sol[-1.03176][r], {r, r0, 1}],Plot[{sol[-1.03176][r], -Sqrt[1 - r^2]}, {r, r0, 1}]}
[![fig2][2]][2]

The author insists that the correct model corresponds to his code. We give a solution for this case. The solution method does not change.

    A1 = 200;
    A2 = 186*10^5;
    A3 = 1/5000;
    a = Rationalize[A2*A3^3];
    H = 1; r0 = 10^-5;
    f[z_, r_] := A1*(z + H) + a*(1/(Sqrt[z^2 + r^2] - 1))^3;
    k = z''[r]/(z'[r]^2 + 1)^(3/2) + z'[r]/(r Sqrt[z'[r]^2 + 1]);
    sol = ParametricNDSolveValue[{k == f[z[r], r], z'[1] == z1, 
        z[1] == -H}, z, {r, r0, 1}, {z1}, WorkingPrecision -> 30];
    {Plot[Evaluate[
        Table[sol[z1][
          r], {z1, .00024816068, .00024816069, .000000000001}]], {r, r0, 
        1}, PlotRange -> All], 
      Plot[Evaluate[
        Table[sol[z1][
          r], {z1, .00024816068, .00024816069, .000000000001}]], {r, 
        r0, .01}, PlotRange -> All]} // Quiet
[![fig3][3]][3]


  [1]: https://i.sstatic.net/bddoj.png
  [2]: https://i.sstatic.net/Jn5uP.png
  [3]: https://i.sstatic.net/UkS7L.png