/bin/bash -c access to unexported variables

Question

I am trying to understand why the following happens: After setting up a variable like var1=test in the current shell, without exporting it (so it will not be available in child processes) and execute /bin/bash -c "echo $var1" right after, I can see the unexported variable. I am aware of the fact that a builtin command like "echo" does not fork() or exec(), so it would have access to the unexported $var1 or about the fact that a SUBSHELL would have access to the unexported variables too.

But in my case, /bin/bash -c "command" is not a SUBSHELL either as /bin/bash -c "echo $BASH_SUBSHELL" would show me "0". And I thought it should behave the same as an external command, so -> fork -> exec and inherit environment variables from the parent. So, for example, I see the following behaivor:

[nimus@localhost ~]$ echo $var1
test
[nimus@localhost ~]$ /bin/bash
[nimus@localhost ~]$ echo $var1
[nimus@localhost ~]$ exit
exit
[nimus@localhost ~]$ /bin/bash -c "echo $var1; echo $BASH_SUBSHELL"
test
0
[nimus@localhost ~]$

So how is /bin/bash -c "command" different then an external command or fork and exec a new whole bash Shell like above?

Thanks!