Looks like the answer is 1011. I'll work out the details, and post the proof (if I'm able to)

Call an island 'lonely' if only 1 bridge is connected to it. Call two islands are 'direcly connected' if there is a bridge between them. We can easily find out that there are at least $2$ lonely islands(since the islands form a tree). Suppose $A_1$ and $A_2$ are two lonely islands. Case 1 $A_1$ and $A_2$ are both connected to an island $B$ Then $A_1,A_2,X$ forms an island group, which means every island group has to include $X$. So starting from $X$, one can get to any island using at most two bridges. That makes the number of lonely islands at least $2+(2023-3)/2=1012$ Case 2 $A_1$ is directly connected to an island $B_1$ and $A_2$ is directly connected to an island $B_2$. Since the islands are connected, $B_1$ and $B_2$ are not lonely. If $B_1$ and $B_2$ are directly connected, as they are also connected to other islands WLOG, suppose $C$ is directly connected to $B_1$, which is different form $A_1,B_2$ No other islands is directly connected to $B_2$ then(otherwise we have two island groups $C,B_1,A_1$ and $D,B_2,A_2$) So in order to have a common island with the island group$B_1,B_2,A_2$, every island group has to contain $B_1$ Similarly to case 1, we have at least $1012$ lonely islands. If $B_1$ and $B_2$ are not directly connected Suppose $B_1$ is directly connected to an island $C_1$ and $B_2$ is directly connected to an island $C_2$. We have two island groups $C_1,B_1,A_1$ and $C_2,B_2,A_2$, they should have 1 common island. So $C_1$ and $C_2$ have to be the same island, we renote them as $C$. That is to say, excluding $A_1$ and $A_2$, $B_1$ and $B_2$ are directly connected to and only directly connected to island $C$ Then every island group has to contain$C$. Therefore we have at least $(2023-1)/2=1011$ lonely islands. In conclusion, there are at least $1011$ lonely islands, the following shows an example. Call the islands $X,Y_1,Y_2,...Y_{1011},Z_1,Z_2,...,Z_{1011}$ Build a bridge between $X$ and $Y_i$, $i=1,2,...,1011$, respectively. Then build a bridge between $Y_i$ and $Z_i$, $i=1,2,...,1011$, respectively. These bridges follows the rules and only the $Z$s are lonely.