A prime question, edited.

Firstly, the prime factors of N must be two less than a prime. This means that the prime factors are congruent to either 3 or 5 modulo 6. Also, if any of the factors were congruent to 1 modulo 6 other than 1, the number would not satisfy the condition. This is because it would be 2 less than a multiple of 3.

Other than 3, suppose that we have 2 other prime factors. Then they must be congruent to 5 modulo 6. However, their product is a factor and thus it would not satisfy the condition. Thus the number is either

(i) equal to $p^{a}$

(ii) equal to $3^{m}p^{n}$

for some prime p>3 and positive integers $a$ , $m$ and $n$ .

However, if $a,n>1$ , $p^{2}$ is a factor of the integer and since $p$ must be congruent to 5 modulo 6, $p^{2}$ is congruent to 1 mod 6 and adding 2 will result in a multiple of 3. Thus the condition will not be satisfied.

To maximise the number of factors, the positive integer must be of type (ii). However, if $m>4$ , $3^{5}=243$ is a factor and adding two will result in a multiple of 5. Thus the maximum possible value of a is 4.

We consider the case when m is 4. We want to prevent any factor from ending with a 3, other than 3, or 2 more than the factor will be a multiple of 5. So p cannot end with a 3. If it ends with a 7, let the factor be x. Since 9 is a factor, 9x which ends with 3 is a factor. Also, if p ends with 9, since 27 is a factor, their product, ending with 3, will also be a factor. Now we are left with 5. However, the positive integer would be $3^{4} \times 5=405$ and 11 is a factor of 407. Thus m cannot be 4.

Note that $3^{3} \times 5=135$ satisfies the condition. Thus the maximum number of factors is 8.