How can I exclude one word with grep?

You can do it using -v (for --invert-match) option of grep as:

grep -v "unwanted_word" file | grep XXXXXXXX

grep -v "unwanted_word" file will filter the lines that have the unwanted_word and grep XXXXXXXX will list only lines with pattern XXXXXXXX.

EDIT:

From your comment it looks like you want to list all lines without the unwanted_word. In that case all you need is:

grep -v 'unwanted_word' file

I understood the question as "How do I match a word but exclude another", for which one solution is two greps in series: First grep finding the wanted "word1", second grep excluding "word2":

grep "word1" | grep -v "word2"

In my case: I need to differentiate between "plot" and "#plot" which grep's "word" option won't do ("#" not being a alphanumerical).

Hope this helps.

If your grep supports Perl regular expression with -P option you can do (if bash; if tcsh you'll need to escape the !):

grep -P '(?!.*unwanted_word)keyword' file

Demo:

$ cat file
foo1
foo2
foo3
foo4
bar
baz

Let us now list all foo except foo3

$ grep -P '(?!.*foo3)foo' file
foo1
foo2
foo4
$ 

The right solution is to use grep -v "word" file, with its awk equivalent:

awk '!/word/' file

However, if you happen to have a more complex situation in which you want, say, XXX to appear and YYY not to appear, then awk comes handy instead of piping several greps:

awk '/XXX/ && !/YYY/' file
#    ^^^^^    ^^^^^^
# I want it      |
#            I don't want it

You can even say something more complex. For example: I want those lines containing either XXX or YYY, but not ZZZ:

awk '(/XXX/ || /YYY/) && !/ZZZ/' file

etc.