#pi# is transcendental, meaning that it is not the root of any polynomial equation with integer coefficients.
Hence #pi^2# is transcendental and irrational too.
If #pi^2# were rational, then it would be the root of an equation of the form:
#ax+b = 0#
for some integers #a# and #b#
Then #pi# would be a root of the equation:
#ax^2+b = 0#
Since #pi# is not the root of any polynomial with integer coefficients, let alone a quadratic, this is not possible.
Further, if #pi^2# was the root of any polynomial equation with integer coefficients then #pi# would be the root of the same equation with each #x# replaced by #x^2#. So since #pi# is transcendental, so is #pi^2#.