I'm going to assume you meant $n$ trials each, just for simplicity. Also, I'm going to assume that the sample success probabilities are computed as averages of your $n$ samples from each distribution.
I'll denote the true probability of success associated with $Y_1$ and $Y_2$ as $p_1$ and $p_2$, respectively. Also, define $\hat{P}_1$ and $\hat{P}_2$ which are the random variables representing our sample proportions, as opposed to $\hat{p}_1$ and $\hat{p}_2$ which are the observed sample proportions. By the Central Limit Theorem we get
$$\frac{\sqrt{n}(\hat{P}_1 - p_1)}{\sqrt{\hat{p}_1(1-\hat{p}_1)}} \xrightarrow{d} \mathcal{N}(0,1)$$
and
$$\frac{\sqrt{n}(\hat{P}_2 - p_2)}{\sqrt{\hat{p}_2(1-\hat{p}_2)}} \xrightarrow{d} \mathcal{N}(0,1)$$.
So for $n$ large, we can approximate the distributions of each of our statistics. Specifically, $\hat{P}_1 \sim \mathcal{N}\left(p_1, \frac{\hat{p}_1(1-\hat{p}_1)}{n}\right)$ and $\hat{P}_2 \sim \mathcal{N}\left(p_2, \frac{\hat{p}_2(1-\hat{p}_2)}{n}\right)$.
We finally get to our test. Our hypotheses are
$$H_0: p_1 - p_2 = 0$$
$$H_1: p_1 - p_2 > 0.$$
Now, we assume we're operating under the null hypothesis, so $p_1 - p_2 = 0$. Under this assumption, we know, approximately,
$$\hat{P}_1 - \hat{P}_2 \sim \mathcal{N}\left(0, \frac{\hat{p}_1(1-\hat{p}_1)}{n} + \frac{\hat{p}_2(1-\hat{p}_2)}{n}\right).$$
We know the distribution of the difference between sample proportions under $H_0$. Given our data, we also can calculate our "observed" difference in sample proportions, $\hat{p}_1 - \hat{p}_2$.
If $P(\hat{P}_1 - \hat{P}_2 > \hat{p}_1 - \hat{p}_2) \leq 0.05$, then we reject $H_0$. Or else, we fail to reject $H_0$.
The interpretation is, if we assume we're operating under the null and we observe a value ($\hat{p}_1 - \hat{p}_2$) that is "inconsistent enough" with the null hypothesis, then we're going to reject $H_0$! This level is precisely $0.05$, because we specified the size of our test.