Well, no idea if this is very efficient, but it seems to solve all situations (the hanoi-driver does not complain) and is based on your code, just generalized.

#! /usr/bin/perl -w
use strict;

my $peg_count  = shift;
my $disk_count = shift;

my @pegs;
{
        my $p = 'A';
        push @pegs , $p++ for 1 .. $peg_count;
}

solve( [ @pegs ] , reverse 1..$disk_count);

sub solve {
        my ($pegs, $first_disk, @rest) = @_;
        my $from = shift @$pegs;
        my $to   = shift @$pegs;
        my $hold = shift @$pegs;
        
        return unless $first_disk;
        
        solve( [ $from, $hold, @$pegs, $to ] => @rest );
        print "$first_disk: $from -> $to\n";
        solve( [ $hold, $to, @$pegs, $from ] => @rest );
}
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Update: Well, it is absolutely NO more efficient than using 3 pegs anyway, but at least it works.