Why is testing “$?” to see if a command succeeded or not, an anti-pattern?

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This is an antipattern because it introduces complexity that wouldn’t exist if you didn’t require the exit status to be recorded at all.

if your_command; then ...

has much less to go wrong than

your_command
if [ "$?" -eq 0 ]; then ...

For examples of things that can go wrong: Think about traps, or even new echo statements added for debugging, modifying $?. It’s not visually obvious to a reader that a separate line running your_command can’t have anything added below it without changing logical flow.

That is:

your_command
echo "Finished running your_command" >&2
if [ "$?" -eq 0 ]; then ...

…is checking the echo, not the actual command.

Thus, in cases where you really do need to deal with exit status in a manner more granular than immediately branching on whether its value is zero, you should collect it on the same line:

# whitelisting a nonzero value for an example of when "if your_command" won't do.
your_command; your_command_retval=$?
echo "Finished running your_command" >&2 ## now, adding more logging won't break the logic.
case $your_command_retval in
  0|2) echo "your_command exited in an acceptable way" >&2;;
  *)   echo "your_command exited in an unacceptable way" >&2;;
esac

Finally: If you enclose your_command inside of an if statement, this marks it as tested, such that your shell won’t consider a nonzero exit status for purposes of set -e or an ERR trap.

Thus:

set -e
your_command
if [ "$?" -eq 0 ]; then ...

…will never (barring a number of corner cases and caveats which plague set -e‘s behavior) reach the if statement with any value of $? other than 0, as the set -e will force an exit in that case. By contrast:

set -e
if your_command; then ...

…marks the exit status of your_command as tested, and so does not consider it cause to force the script to exit per set -e.

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