Long story short: this is fine, no monkey business necessary, as long as $|z|$ is sufficiently small.
Long story long:
This will certainly work for $z$ with a sufficiently small modulus. It comes down to the following:
Let $\|\cdot\|$ be any sub-multiplicative matrix-norm. If $\sum a_n z^n$ is a power series with radius of convergence $R>0$, then $\sum a_n X^n$ will converge for the matrix $X$ whenever $\|X\| < R$ (or equivalently, whenever $\rho(X) < R$).
Now, your square-root series has a radius of convergence $1$. So, as long as $\|zX\| < 1$ for some sub-multiplicative matrix norm, the sum $\sum a_n (zX)^n$ will converge.
You can then confirm (using the Cauchy product) that $(\sum a_n z^n)^2 = 1 + z$. Since such manipulation is allowed when the sum is absolutely convergent, you'll find that $(\sum a_n (zX)^n)^2 = I + zX$, which is to say that this series will give you some square root of the matrix $I + zX$.
Note that if $z = 1$, the condition $|X_{ij}| < 1$ is insufficient because the max-norm is not sub-multiplicative.