Solve`DirInf[] — Meaningful value or just a bug?

Question

Bug introduced in 4.1 or earlier and persisting through 12.2.0. or later

To the problem below, I get four independent, incomplete solutions, three in terms of Solve`DirInf[].  Since DirInf is not in the System`  context, I assume this behavior is a bug (or not?).  But I was wondering if it made any sense as a solution.
Solve[Flatten@{
   NestList[
     D[#, x] &, (
      Hypergeometric2F1[1/2, 2/3, 5/3, (8 I y[x]^(3/2))/(3 C[1])]^2 y[
        x]^2 (1 - (8 I y[x]^(3/2))/(3 C[1])))/(
      C[1] - 8/3 I y[x]^(3/2)) == (x + C[2])^2, 1] /. {x -> 0},
   y[0] == 1, y'[0] == 0},
 {C[1], C[2]}, {y[0], y'[0]}]

Solve::ifun :  Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
Solve::svars :  Equations may not give solutions for all "solve" variables. >>
{{C[1] -> -((8 I)/3)}, {C[1] -> (8 I)/3},
 {C[1] -> -(8/(3 Sqrt[-1 + 2 Solve`DirInf[] - Solve`DirInf[]^2]))},
 {C[1] -> 8/(3 Sqrt[-1 + 2 Solve`DirInf[] - Solve`DirInf[]^2])},
 {C[2] -> 0}}

BTW, Reduce fails on the system.
FWIW, the problem arises from the following IVP:
y''[x]^2 == -4 y[x] && y[0] == 1 && y'[0] == 0

Akku14 · Answer

I don't think, it'a a bug. Two ways to get the result you want.
First omit the comand to eliminate {y[0], y'[0]}  in Solve. Since you explicitly set them Equal 1 and 0 and later want to eliminate it, Solve gets confused.
eqs = Flatten@{NestList[
 D[#, x] &, (Hypergeometric2F1[1/2, 2/3, 
       5/3, (8 I y[x]^(3/2))/(3 C[1])]^2 y[
       x]^2 (1 - (8 I y[x]^(3/2))/(3 C[1])))/(C[1] - 
     8/3 I y[x]^(3/2)) == (x + C[2])^2, 1] /. {x -> 0}, y[0] == 1,
y'[0] == 0}

Solve[eqs, {C[1], C[2]}]

(*   {{C[1] -> (8 I)/(
3 InverseFunction[Hypergeometric2F1, 4, 4][1/2, 2/3, 5/3, 0]), 
C[2] -> 0}}   *)

Second, set the y with Rule to wanted values to get the same result
Solve[Flatten@
  Evaluate@{NestList[
   D[#, x] &, (Hypergeometric2F1[1/2, 2/3, 
         5/3, (8 I y[x]^(3/2))/(3 C[1])]^2 y[
         x]^2 (1 - (8 I y[x]^(3/2))/(3 C[1])))/(C[1] - 
       8/3 I y[x]^(3/2)) == (x + C[2])^2, 1] /. 
  x -> 0 /. {y[0] -> 1, y'[0] -> 0}}, {C[1], C[2]}, {y[0], y'[0]}
]

Edit
I am using version "8.0 for Microsoft Windows (32-bit) (December 9, 2010)"
@MichaelE2, you are right. Solve should be able to give solution for (but does not)
Solve[{c1 == y1, c2 == y2, y1 == 1, y2 == 0}, {c1, c2}]

Don't know why my first example works this time without command to eliminate the y.
Second, if you accept infinity as solution, C[1] -> +/- infinity together with C[2]==0 are solutions.
eqs2 = eqs /. {y[0] -> 1, y'[0] -> 0}

Plot[Evaluate@Through[{Re, Im}[eqs2[[1, 1]] /. C[1] -> c1]], {c1, -5, 
   5}, PlotStyle -> {Blue, Red}]

Limit[eqs2[[1, 1]], C[1] -> -[Infinity]]

(*   0   *)

eqs /. {C[1] -> -[Infinity], C[2] -> 0}

(*   {True, True, y[0] == 1, Derivative[1][y][0] == 0}   
*)

Solve`DirInf[] — Meaningful value or just a bug?

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