Picking kr as the argument, take a look at J(kr) and do the differentiations on
d^2/dr^2 J(kr) + (1/r) d/dr(J(kr).
Take out a factor of k for each derivative, e.g.
where the prime always denotes differentiation by the entire argument.
So for kr = z, then formally
Proceeding like this (and not forgetting that 1/r is going to become k/z) you can match the top equation for nu = 0, so the solution really is I0(kr). All the In(kr) work similarly, including I1(kr).
You can also do a more dimensional analysis style derivation. Every term in the desired equation
d^2/dr^2 J + (1/r) d/dr J - k^2 J = 0
has dimension length^-2. Divide through by k^2 and combine k with r
d^2/d(kr)^2 J + (1/kr) d/d(kr) J - J = 0
This equation is dimensionless and matches the correct differential equation for I0, on the condition that you use the dimensionless quantity kr as the argument for I0.