How does bash <command-argument> work?

About bash For a new _tty_ when is executed the echo $SHLVL command it displays 1 as expected. Now, if in the same _tty_ is executed the bash command and later again the echo $SHLVL command it displays 2. Is mandatory use the exit command to exit of course. Furthermore I did do realize that each bash has its own _history command_ and the user is interacting or has access in the current bash. So far, after to did do a research it seems _it is a kind of subshell_ (correct me if I am wrong), it because mostly a subshell is created through the () approach instead. Just as playing I executed the bash cat /etc/os-release command and nothing is printed. Therefore *being curious* **Question** * How does bash work? As _extra question_: * Under what circumstances the bash approach would be mandatory to be applied? **Observation** In the answer was indicated to expect an error message. Well, the error mentioned is correct (tested on Ubuntu and Fedora) but in my case the bash cat /etc/os-release command was applied as an argument to create a docker container based on Linux and none error was shown as indicated from the beginning and it was the reason to create this post. Well it is other history