Let's solve this first for the case where $X$ and $Y$ are iid standard Normal variables. Because $X$ and $Y$ are identically distributed, the $Z_i$ are exchangeable and therefore also must be identically distributed.

Let their common variance be $\sigma^2$ and suppose for a moment it is finite. Since $Z_1+Z_2=1$ is constant it has zero variance, allowing us to compute

$$0 =\operatorname{Var}(Z_1+Z_2) = 2\sigma^2 + 2 \operatorname{Cov}(Z_1,Z_2),$$

thereby deducing

$$\operatorname{Cov}(Z_1,Z_2) = -\sigma^2.$$

But, writing $\mu=E[Z_1] =E[Z_2]$ (whose finiteness is assured by the assumed finiteness of $\sigma^2$), algebra shows us

$$\eqalign{ \sigma^2 = -\operatorname{Cov}(Z_1,Z_2) &= -E[Z_1 Z_2] + \mu^2 \\ &= -E\left[\frac{XY}{(X+Y)^2}\right] + \mu^2\\ &= -\frac{1}{4} E\left[\frac{(X+Y)^2 - (X-Y)^2}{(X+Y)^2}\right] + \mu^2\\ &= \frac{1}{4}\left(-1 + E\left[\frac{(X-Y)^2}{(X+Y)^2}\right]\right) + \mu^2. }$$

The numerator and denominator in that final fraction are independent because $(X-Y, X+Y)$ are jointly Normally distributed with zero covariance. Since a multiple of the denominator $(X+Y)^2$ must therefore have a $\chi^2(1)$ distribution, and the density of that distribution in a neighborhood of $0$ is positive, the ratio must have infinite variance.

We are compelled to reject the original assumption that $\sigma^2$ is finite; and there is the answer: $\operatorname{Var}(Z_1) = \operatorname{Var}(Z_1) = \infty.$

It is straightforward to generalize these arguments to arbitrary independent Normal variables--the algebra gets a little messier, but the same principle applies: the ratios are random variables with infinite variances.

Let's solve this first for the case where $X$ and $Y$ are iid standard Normal variables. (The question appears to assume $X$ and $Y$ are identically distributed anyway, suggesting this is a good starting point.)

Because $X$ and $Y$ are identically distributed, the $Z_i$ are exchangeable and therefore also must be identically distributed. Let their common variance be $\sigma^2$ and suppose for a moment it is finite. Since $Z_1+Z_2=1$ is constant it has zero variance, allowing us to compute

$$0 =\operatorname{Var}(Z_1+Z_2) = 2\sigma^2 + 2 \operatorname{Cov}(Z_1,Z_2),$$

thereby deducing

$$\operatorname{Cov}(Z_1,Z_2) = -\sigma^2.$$

But, writing $\mu=E[Z_1] =E[Z_2]$ (whose finiteness is assured by the assumed finiteness of $\sigma^2$), algebra shows us

$$\eqalign{ \sigma^2 = -\operatorname{Cov}(Z_1,Z_2) &= -E[Z_1 Z_2] + \mu^2 \\ &= -E\left[\frac{XY}{(X+Y)^2}\right] + \mu^2\\ &= -\frac{1}{4} E\left[\frac{(X+Y)^2 - (X-Y)^2}{(X+Y)^2}\right] + \mu^2\\ &= \frac{1}{4}\left(-1 + E\left[\frac{(X-Y)^2}{(X+Y)^2}\right]\right) + \mu^2. }$$

The numerator and denominator in that final fraction are independent because $(X-Y, X+Y)$ are jointly Normally distributed with zero covariance. Since a multiple of the denominator $(X+Y)^2$ must therefore have a $\chi^2(1)$ distribution, and the density of that distribution in a neighborhood of $0$ is positive, the ratio must have infinite variance.

We are compelled to reject the original assumption that $\sigma^2$ is finite; and there is the answer: $\operatorname{Var}(Z_1) = \operatorname{Var}(Z_2) = \infty.$

It is straightforward to generalize these arguments to arbitrary independent Normal variables--the algebra gets a little messier, but the same principle applies: the ratios are random variables with infinite variances.

BTW, because the $Z_i$ sum to unity their means must either sum to unity or be undefined, whence $\mu=1/2$ (returning to the original simplified setting) or else $\mu$ is undefined. This is a very general result, holding for any bivariate random variable $(X,Y)$ where $X$ and $Y$ have identical expectations $\mu.$