A Numpad's Knight Numbers

Python 2, 52 bytes

f=lambda n:n<6or`n%100`in'18349276167294381'*f(n/10)

Checks that any two consecutive digits are in the string '18349276167294381'. To get consecutive digits, rather than doing zip(`n`,`n`[1:]), the function repeatedly checks the last two digits and removes the last digit.

Jelly, 19 15 14 bytes

Doȷ’d3ạ2\P€=2P

Try it online! or verify all test cases.

How it works

Doȷ’d3ạ2\P€=2P  Main link. Argument: n (integer)

D               Convert n to base 10 (digit array).
  ȷ             Yield 1000.
 o              Logical OR. This replaces each 0 with 1000.
   ’            Decrement each digit.
    d3          Divmod; replace each digit k with [k:3, k%3].
      ạ2\       Pairwise reduce by absolute difference.
                For each pair of adjacent digits [i, j], this computes
                [abs(i:3 - j:3), abs(i%3 - j%3)].
         P€     Compute the product of each result.
                n is a Numpad's Knight Number iff all products yield 2.
           =2   Compare each product with 2.
             P  Multiply the resulting Booleans.

Retina, 58 40 bytes

Thanks to Sp3000 for suggesting this idea:

M&!`..
O%`.
A`16|18|27|29|34|38|49|67
^$

Try it online! (Slightly modified to run the entire test suite at once.)

Prints 1 for truthy and 0 for falsy results.

Explanation

M&!`..

Find all overlapping matches of .., i.e. all consecutive pairs of digits, and join them with linefeeds.

O%`.

Sort the digits in each line, so that we only need to check half as many pairs.

A`16|18|27|29|34|38|49|67

Remove all lines which correspond to a valid move.

^$

Count the matches of this regex. That is, if all lines were removed, this matches the resulting empty string once, otherwise it fails to match and gives zero instead.