No

Suppose such a set exists, and let $p$ be the biggest prime of this set. Now, $2^2 + 3^2 +...+ n^2=1^2 + 2^2 + 3^2 +...+ n^2 = \frac{n(n+1)(2n+1)}{6} -1$. Now let $n$ equal to the product of product of all primes up till $p$ multiplied by 6 $=> n=2 \cdot 3 \cdot 5 \cdot ... \cdot p \cdot (6)$. Now, clearly, all primes up till $p$ will divide $\frac{n(n+1)(2n+1)}{6}$, thus no prime up till $p$ will divide $\frac{n(n+1)(2n+1)}{6}-1$, hence $p$ cannot be the largest possible prime, hence contradiction. Note: Yes, I know it can be easily optimized far better, but this works as well and it is easier to deal with.