In case you want to compare files with the same name (that exist in both lists/directories), you could compute the intersection of both lists, and then iterate over the resulting list of files names, simply prepending the appropriate paths... Something like:

my %seen;
$seen{$_}++ for @return, @remoteFilelist;
my @files_in_both_lists = grep $seen{$_} > 1, keys %seen;

for my $fname (@files_in_both_lists) {
    if (compare_text("$path1/$fname", "$path2/$fname") == 0) {
        #...
    }
}
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Otherwise (if you want to compare every file in list 1 with every file in list 2), I would compute checksums (e.g. MD5) for all files, and use the checksums as keys in a hash, with a list of filenames as the associated value. Those entries with more than one file in that list will indicate identical files...

Update: sample code for the latter approach:

#!/usr/bin/perl

use strict;
use warnings;
use Digest::MD5;

my @allfiles = ...;   # your file lists merged (including paths)

my %by_md5;

for my $file (@allfiles) {

    open my $fh, "<", $file or die "Couldn't open '$file': $!";
    binmode $fh;
    
    my $md5 = Digest::MD5->new();
    $md5->addfile($fh);
    my $digest = $md5->hexdigest();  # or ->digest() -- hexdigest is j
+ust more "dumping-friendly"...

    push @{ $by_md5{$digest} }, $file;
}

for my $digest (grep @{$by_md5{$_}} > 1, keys %by_md5) {
    print "duplicates: @{ $by_md5{$digest} }\n";
}
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(In case you're paranoid (and worry about the very unlikely case of a digest collision), you can always do a byte-for-byte comparison of the files with the same digest...(those reported as duplicates with the above snippet))