There is a confusion between the power and the adequation level.
Let's assume that we test
$$
H_0:\theta = 1 \qquad\text{v.s.}\qquad H_1: \theta = 10 \, .
$$
The level $\alpha$ is chosen so as to set a bound for the first-type error, which is defined as the probability of rejecting incorrectly the null hypothesis $H_0$. In other words, $\alpha = \Bbb P(\text{reject } H_0 | H_0\text{ is true})$. A typical choice for the level is $\alpha = 0.05$. Now, we can define the adequation level-rejection region $W_\alpha$ such that
$$
\alpha = \Bbb P_{\theta=1}((X_1,\dots ,X_n)\in W_\alpha) \, .
$$
If the observation $(x_1,\dots ,x_n)$ belongs to $W_\alpha$, then we reject $H_0$.
Similarly, we define the second-type error $\beta$ as the probability of accepting incorrectly the null hypothesis $H_0$. In other words,
\begin{aligned}
\beta &= \Bbb P(\text{accept } H_0 | H_1\text{ is true})\\
&=\Bbb P_{\theta=10}((X_1,\dots ,X_n)\in W_\alpha^c)\, ,
\end{aligned}
where $W_\alpha^c = \Bbb R^n \setminus W_\alpha$. The power of the test is $1-\beta$, i.e., the probability of rejecting correctly the null hypothesis $H_0$. In other words, the power corresponds to
\begin{aligned}
1-\beta &= \Bbb P(\text{reject } H_0 | H_1\text{ is true})\\
&= \Bbb P_{\theta=10}((X_1,\dots ,X_n)\in W_\alpha) \, .
\end{aligned}
If we have to choose among several tests, we choose one with the largest possible statistical power.
This may be summarized in the following table of probabilities $\Bbb P(\text{decision|fact})$:
$$
\begin{array}{c|cc}
\text{decision\fact} & H_0 \text{ is true} & H_1 \text{ is true} \\
\hline
\text{reject } H_0 & \alpha & 1-\beta \\
\text{accept } H_0 & 1-\alpha & \beta
\end{array}
$$
