First I rationalized your equations. Then I noticed that the equation for x[t] separates and is very easy. Replacing this solution into the equation for z[t] I saw that this becomes singular for a value of t between 0 and 1:
F = Derivative[3][x][
t]^2 + ((-3*(-0.75 + x[t])*
Derivative[1][x][
t]^3)/(0.48999999999999994 - (-0.75 + x[t])^2)^(3/
2) - (3*(-0.75 + x[t])^3*
Derivative[1][x][
t]^3)/(0.48999999999999994 - (-0.75 + x[t])^2)^(5/
2) - (3*Derivative[1][x][t]*Derivative[2][x][t])/
Sqrt[0.48999999999999994 - (-0.75 + x[t])^2] - (3*(-0.75 +
x[t])^2*Derivative[1][x][t]*
Derivative[2][x][
t])/(0.48999999999999994 - (-0.75 + x[t])^2)^(3/2) +
6*Derivative[1][z][t]*
Derivative[2][z][t] - ((-0.75 + x[t])*Derivative[3][x][t])/
Sqrt[0.48999999999999994 - (-0.75 + x[t])^2] +
2*z[t]*Derivative[3][z][t])^2 //
Rationalize // Simplify;
bdryX = {x[0] == 0, x[1] == 1, x'[0] == 0,
x''[0] == 0, x'[1] == 0, x''[1] == 0};
bdryZ = {z[0] == 0.3162, z[1] == 0.5884,
z''[0] == 0, z'[1] == 0, z''[1] == 0} // Rationalize;
eqs0 = Solve[{D[D[F, z'''[t]], {t, 3}] + D[D[F, z'[t]], {t, 1}] ==
D[D[F, z''[t]], {t, 2}] + D[F, z[t]],
D[D[F, x'''[t]], {t, 3}] + D[D[F, x'[t]], {t, 1}] ==
D[D[F, x''[t]], {t, 2}] + D[F, x[t]]} // Simplify,
D[{x[t], z[t]}, {t, 6}]][[1]] /.
Rule -> Equal // Simplify;
eqx = eqs0[[1]]
solx = DSolve[Join[{eqx}, bdryX], x, t][[1]]
eqz = eqs0[[2]] /. solx
eqz /. Solve[Denominator[eqz[[2]]][[1, 1]] == 0 &&
0 < t < 1][[1]] // Simplify
