Answer: 270

Small Hint:

Handle $3^3,$ $5^5,$ $7^7$ separately. Since $149 - 2 = 147 = 3 \cdot 7^2,$ lifting the exponent applies directly at $3$ and $7.$

Big Hint:

At $5,$ first $4^n \equiv 2^n \pmod 5$ requires $4$ to divide $n.$ Then $v_5(149^4 - 2^4) = 1,$ so lifting the exponent forces $5^4$ to divide $\frac{n}{4}.$

Solution:

Work prime by prime. Since $149 - 2 = 147 = 3 \cdot 7^2,$ the lifting-the-exponent lemma gives $v_3(149^n - 2^n)$ $= v_3(147) + v_3(n)$ $= 1 + v_3(n)$ and $v_7(149^n - 2^n) = 2 + v_7(n)$ for every positive integer $n.$ Requiring at least $3$ and $7$ forces $n$ to be divisible by $3^2$ and by $7^5.$

For $5$ we first need $149^n \equiv 2^n \pmod 5,$ i.e. $4^n \equiv 2^n,$ i.e. $2^n \equiv 1 \pmod 5,$ which requires $4$ to divide $n.$ Write $n = 4k.$ In $149^4 - 2^4$ $= (149 - 2)(149 + 2)(149^2 + 4),$ only the last factor is divisible by $5,$ and only once, since $149^2 + 4 = 22205 = 5 \cdot 4441.$ Lifting the exponent from the base $149^4, 2^4$ gives $v_5(149^n - 2^n) = 1 + v_5(k),$ so $5^4$ divides $k,$ i.e. $4 \cdot 5^4$ divides $n.$

The least valid $n$ is $2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5,$ which has $(2+1)(2+1)(4+1)(5+1)$ $= 270$ positive divisors.