It is true that if $N > 60$, then $2^{N} - 1$ has a prime factor $> 2500$.

Here's another approach. First, observe that every prime factor of $2^{p} - 1$ is $\equiv 1 \pmod{p}$. This implies that if $2^{N} - 1$ has all prime factors $\leq 2500$, then all prime factors of $N$ are $< 2500$. Checking these primes, we see that $2^{p} - 1$ has a prime factor $> 2500$ unless $p = 2, 3, 5, 7, 11$ or $29$. Factoring $2^{p^{k}} - 1$ for $p \in \{ 2, 3, 5, 7, 11, 29 \}$ shows that the exponent on such $p$s cannot be larger than $4$, $2$, $2$, $2$, $1$ and $1$ in the respective cases. Checking this finite collection of numbers shows that $N = 60$ is the largest possible.

It is true that if $N > 60$, then $2^{N} - 1$ has a prime factor $> 2500$.

Here's another approach. First, observe that every prime factor of $2^{p} - 1$ is $\equiv 1 \pmod{p}$. Combining this with the observation that if $a | b$, then $2^{a} - 1 | 2^{b} - 1$, we see that if $2^{N} - 1$ has all prime factors $\leq 2500$, then all prime factors of $N$ are $< 2500$. Checking these primes, we see that $2^{p} - 1$ has a prime factor $> 2500$ unless $p = 2, 3, 5, 7, 11$ or $29$. Hence if all the prime factors of $2^{N} - 1$ are less than $2500$, then all prime divisors of $N$ are in the set $S = \{ 2, 3, 5, 7, 11, 29 \}$.

We find that $65537$ is a prime factor of $2^{32} - 1$ and this means that $N$ cannot be a multiple of $32$ if $2^{N} - 1$ has all prime divisors $< 2500$. Similar arguments show that $N$ cannot be a multiple of $3^{3}$, $5^{3}$, $7^{3}$, $11^{2}$ or $29^{2}$. This implies that $N$ divides $56271600$, and checking all such divisors, we see that $N = 60$ is the largest possible.