Do you know that...

Yes, it's documented in perlapi. I didn't assume that would be a concern for the OP.

The old pure-Perl version of Scalar::Util has a pure-Perl version of looks_like_number, which includes the following code:

return 1
  if ( $] >= 5.008    and /^(Inf(inity)?|NaN)$/i ) 
  or ( $] >= 5.006001 and /^Inf$/i               );
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So if we wanted to eliminate that behavior, we could either copy/paste and modify the entire pure-Perl version from an old version of Scalar::Util (but it has issues with '0 but true'), or use the modern version of Scalar::Util::looks_like_number like this:

sub looks_like_finite_number {
  my $candidate = shift;
  return 0 if ! looks_like_number( $candidate );
  return 0
    if ( $] >= 5.008    and $candidate =~ /^(Inf(inity)?|NaN)$/i) 
    or ( $] >= 5.006001 and $candidate =~ /^Inf$/i );
  return 1;
}
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Or, in perlfaq4 there is a series of regexes given for determining if one is looking at a number under the section, "How do I determine whether a scalar is a number/whole/integer/float? It's not pretty, and that FAQ answer has even changed over the years; it used to use a bunch of "if" statements. Now it uses given/when. *sigh*