Let us say that $H_i \in \{\text{aids}, \text{healthy}\}$ and $A \in \{\text{positive},\text{ negative}\}$. Next, from the problem statement we know that
\begin{align}
&\mathbb{P}(A = \text{positive} \mid H_i = \text{aids}) = 0.95\\
&\mathbb{P}(A = \text{negative} \mid H_i = \text{healthy}) = 0.95 \\
& \mathbb{P}(H_i = \text{aids}) = 0.05 \tag 1
\end{align}
and we are asked for the value of $\mathbb{P}(H_i = \text{aids} \mid A = \text{positive})$. Exactly as you proposed, Bayes' rule is what has to be used here. In particular we use
\begin{equation*}
\mathbb{P}(H_i = \text{aids} \mid A = \text{positive})= \frac{\mathbb{P}(A = \text{positive} \mid H_i = \text{aids}) \mathbb{P}(H_i = \text{aids}) }{\mathbb{P}(A = \text{positive})} \tag 2
\end{equation*}
From (1) we have all values in the left part of (2) except for the evidence $\mathbb{P}(A = \text{positive})$, which can be computed marginalizing over $H_i$ in the joint probability $\mathbb{P}(A = \text{positive},H_i)$, as highlighted in (3).
\begin{align}
\mathbb{P}(A = \text{positive}) = {} & \sum_{H_i} \mathbb{P}(A = \text{positive},H_i) = \sum_{H_i} \mathbb{P}(A = \text{positive}\mid H_i) \mathbb{P}(H_i)\\
= {} & \mathbb{P}(A = \text{positive}\mid H_i = \text{aids}) \mathbb{P}(H_i = \text{aids}) \\
& {} + \mathbb{P}(A = \text{positive}\mid H_i = \text{healthy}) \mathbb{P}(H_i = \text{healthy}) \tag 3
\end{align}
Note that in (3) two new probabilities appear, which however can easily be computed using the complementary probabilities of the given data in (1) as
\begin{equation}
\begin{split}
&\mathbb{P}(A = \text{positive} \mid H_i = \text{healthy}) = 1-\mathbb{P}(A = \text{negative} \mid H_i = \text{healthy}) = 0.05\\
& \mathbb{P}(H_i = \text{healthy}) = 1- \mathbb{P}(H_i = \text{aids}) = 0.95
\end{split} \tag 4
\end{equation}
Substituting the results from (4) in (3) we obtain that $\mathbb{P}(A = \text{positive}) = 0.095$. Finally, using this results we compute (2) as
\begin{equation*}
\mathbb{P}(H_i = \text{aids} \mid A = \text{positive})= \frac{0.95\cdot0.05}{0.095} = 0.5
\end{equation*}
The test turns out to be not really trustful. This is basically due to high population percentage having aids.
P.S.
I recommend you the reading of these two articles, they are very interesting and can give you rich insights.
