I coudl work with something more like - but I guess I run into the same problem.
eq1 = 1 == b*c^m1 + a
eq2 = 3/4 == b*c^(m) + a
eq3 = 0 == b*c^(0) + a
Reduce[{eq1, eq2, eq3}, {a, b, cm}] // ToRadicals
which if parametrise for m and m1 gets me workable coefficients form a,b, and c. So i wonder if there's a way that if with some assumption could lead me to a symobolic closed form.
with m1 = 100 and m=25
Solve[{eq1, eq2, eq3}, {a, b, c}, Reals] // ToRadicals
{{a -> 1 - ((-3 - 1/(3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(
1/3)) (-2 - 1/(3 - 2 Sqrt[2])^(2/3) + 3/(3 - 2 Sqrt[2])^(
1/3) + 3 (3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(2/3))^4)/
6912, b ->
3/16 (-3 - 1/(3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(1/3)),
c -> 1/(6/(-2 - 1/(3 - 2 Sqrt[2])^(2/3) + 3/(3 - 2 Sqrt[2])^(1/3) +
3 (3 - 2 Sqrt[2])^(1/3) - (3 - 2 Sqrt[2])^(2/3)))^(1/25)}}