Let me attempt to turn my comment into an answer. The paper
Balog, Antal; Wooley, Trevor D., On strings of consecutive integers with no large prime factors, J. Aust. Math. Soc., Ser. A 64, No. 2, 266-276 (1998). ZBL0942.11041.
contains (a stronger version of) the following result: For each positive integer $k$ and each $\epsilon > 0$, there are infinitely many $n$ where none of $n+1, \dots, n+k$ has a prime factor exceeding $n^\epsilon$.
Here's the basicseed of the idea. For 3 consecutive numbers, one can look at $x^m-1$, $x^m$, $x^m+1$, where $m$ is odd and has many small prime factors. Certainly $x^m$ has only small prime factors relative to its size. And $x^m-1$ and $x^m+1$ factor into cyclotomic polynomials of degree at most $\phi(m)$. So all the prime factors should be of size at most about $x^{\phi(m)}=(x^m)^{\phi(m)/m}$, and $\phi(m)/m$ can be made as small as one wants (by pumping $m$ full of small odd primes). Ultimately this comes down to the Euler's result that $\sum 1/p$ diverges.
It might seem we are stuck, because (e.g.) $x^m-2$ will not have such a nice factorization. But we are free to choose $x$. Suppose we try $x=2^a$. Then $x^m-2 = 2(2^{am-1}-1)$. If we pick $a$ appropriately, we can make $am-1$ also chock full of small primes (we can't get any of the primes appearing in $m$, but we can still put in enough to make sure the reciprocal sum of the primes dividing $am-1$ is large). Then $2^{am-1}-1$ will have only small prime factors relative to its size, for reasons like the last paragraph. And one can play similar games to extend the string to any length, but extracting a quantitative result (as in the Balog--Wooley paper) gets a bit messy.