Assuming (as you do in a comment) that the $\theta_i$ have the same sign and $\theta_2$ is nonzero, the null hypothesis is algebraically equivalent to
$$H_0: \theta_1 = \theta_2$$
while the two alternative hypotheses are equivalent to
$$H_A:\theta_1\ne\theta_2\quad\text{and}\quad H_A^\prime:\theta_1\lt\theta_2\lt 0\text{ or } 0\lt \theta_2\lt\theta_1.$$
Since all the $\epsilon_1$ are independent of the $\epsilon_2,$ all the $Y_1$ responses are independent of the $Y_2$ responses and so (presuming the estimates are separately computed, one set for the $Y_1$ data and another set for the $Y_2$ data) the parameter estimates $\hat\theta_i$ are independent.
How you proceed depends on circumstances. To sketch the general approach, let's suppose you would use a $t$ test or $Z$ test in either regression alone. This means the combined information of assumptions and data is strong enough to suggest the sampling distributions of the $\hat\theta_i$ are approximately Normal with estimated variances $\sigma_i^2$ respectively. Consequently, the test statistic
$$\hat\theta = \hat\theta_1 - \hat\theta_2$$
is approximately Normal with sampling variance
$$\operatorname{Var}(\hat\theta) = \operatorname{Var}(\hat\theta_1) + \operatorname{Var}(\hat\theta_2) = \sigma_1^2 + \sigma_2^2.$$
You would therefore refer the test statistic
$$Z = \frac{\hat\theta_2 - \hat\theta_2}{\sqrt{\sigma_1^2 + \sigma_2^2}}$$
to the Standard Normal distribution with distribution function as $\Phi.$ The critical region for testing $H_0$ against $H_A$ for a test with confidence $1-\alpha$ therefore is
$$\mathcal{C}(\alpha) = (-\infty, \Phi^{-1}(\alpha/2)]\ \cup\ [\Phi^{-1}(1-\alpha/2),\infty).$$
Comments
Unless you're pretty sure what the common sign of the $\theta_i$ is, $\mathcal{C}(\alpha)$ would be the critical region for $H_A^\prime,$ too.
When both the coefficient estimates $\hat\theta_i$ indicate they are significantly different from zero with the same sign, for testing $H_A^\prime$ you could instead replace $\mathcal C(\alpha)$ by the one-sided region
$$\mathcal C^\prime(\alpha) = (-\infty, \Phi^{-1}(\alpha)]$$
(for a common negative estimate) or its negative (for a common positive estimate), thereby achieving greater power. Although the confidence of this test would be slightly less than $1-\alpha,$ the difference shouldn't matter. A quick simulation study adapted to data like yours would remove any doubts.
When one (or both) of the datasets is "small" (around $20$ or less, approximately), consider a Welch t-test instead of a Z test.
