You can use the first two equations to get an algebraic equation for z[k]
:
(* x[k+1] - y[k+1] then compare to third equation *)
zksol = Simplify[(-(Cos[(k π)/2]/k!) + z[k])/(1 + k) - (1/k! +
z[k])/(1 + k) /. k -> (k - 2), Assumptions -> {k ∈ Integers}]
(* -((2 Sin[(k π)/4]^2)/((-1 + k) (-2 + k)!)) *)
Then you can solve a smaller system
{xksol[k_], yksol[k_]} = {x[k], y[k]} /.
First@RSolve[{x[1 + k] == (-(Cos[(k π)/2]/k!) + zksol)/(1 + k),
y[1 + k] == (1/k! + zksol)/(1 + k)}, {x[k], y[k]}, k]
(* {(2 Cos[1/2 (-1 + k) π] (-3 + k)! -
k Cos[1/2 (-1 + k) π] (-3 + k)! - (-1 + k)! +
Cos[1/2 (-1 + k) π] (-1 + k)!)/((-2 k +
k^2) (-3 + k)! (-1 + k)!), (-2 (-3 + k)! +
k (-3 + k)! - (-1 + k)! +
Cos[1/2 (-1 + k) π] (-1 + k)!)/((-2 k +
k^2) (-3 + k)! (-1 + k)!)} *)
Check:
Simplify[{x[1 + k] == (-(Cos[(k π)/2]/k!) + z[k])/(1 + k),
y[1 + k] == (1/k! + z[k])/(1 + k)} /.
{x[p_] -> xksol[p], y[p_] -> yksol[p], z[k] -> zksol}]
(* {True, True} *)
Notice there is no freedom left for initial conditions.