Your 2nd test would be better as

if (ref $test[1] and exists $test[1][1]) {}
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It occurs to me that a sub that returns the passed value iff it is a ref might provide some shorthand for this idiom:

sub iff_ref {
  return ref $_[0] ? $_[0] : undef;
}
# Then your test becomes (if exists doesn't mind the notation)
if (exists iff_ref($test[1])->[1]) {}
# or it might have to be
if (exists ${iff_ref($test[1])}[1]) {}
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I don't have perl5 where I am today, so this is absolutely untested.