I will try to take this a little further.
This gets a clean answer for c==7, z==7.
Solve[(x + y)^z ==
x^c + Sum[Binomial[c, n] x^(c - n) y^(n), {n, 1, c}], z, Reals]
Solve[(x + 1)^z == Sum[Binomial[c, n] x^(n), {n, 0, c}], z, Reals]
The result of this is more complex, but is found by Mathematica: z is a long equation when c==7, mathematica warns inverse functions may not show all solutions...
Solve[(x + y)^z ==
x^c + Sum[Binomial[c, n] x^(c - n) y^(n), {n, 1, c}], z, Complexes]
Solve[(x + 1)^z == Sum[Binomial[c, n] x^(n), {n, 0, c}], z, Complexes]
I'm sure you may have a complex binomial theorem - and one does not immediately jump to my mind other than DeMoirve's, (r(Cos[th]+I Sin[th])^n == r^n(Cos[n th]+I Sin[n th]). (which worker together with euler imaginary identities as well). This shows if your variable "x" can be expressed in terms f[x]==Cos[x] by parameterization (which seems reasonable), then there are binomial theorems that apply that do not apply to Real numbers.
You are perhaps far ahead of me on study of the Complex Plane so I won't continue. But different theorems may be needed depending on the number of variables, whether the exponent is limited in sign or itself or if the exponent may be real and and itself complex. What the objective (in learning) is makes a difference.
https://mathworld.wolfram.com/BinomialTheorem.html