1. **State the problem:** We need to find the number of pairs of positive integers $(n,m)$ such that the sum of factorials from $1!$ to $n!$ is a perfect square, i.e., $$1! + 2! + 3! + \cdots + n! = m^2.$$ 2. **Calculate the sum for small values of $n$: ** Factorials grow very fast, so start by computing the sum for small $n$ values and check if the result is a perfect square. - $n=1: 1! = 1$ which is $1^2$, a perfect square. - $n=2: 1! + 2! = 1 + 2 = 3$, not a perfect square. - $n=3: 1! + 2! + 3! = 1 + 2 + 6 = 9 = 3^2$, a perfect square. - $n=4: 1! + 2! + 3! + 4! = 9 + 24 = 33$, not a perfect square. - $n=5: 33 + 120 = 153$, not a perfect square. - $n=6: 153 + 720 = 873$, not a perfect square. - $n=7: 873 + 5040 = 5913$, not a perfect square. - $n=8: 5913 + 40320 = 46233$, not a perfect square. - $n=9: 46233 + 362880 = 409113$, not a perfect square. - $n=10: 409113 + 3628800 = 4037913$, not a perfect square. 3. **Analyze the pattern:** Because factorials grow very fast and sum quickly becomes large and not a perfect square, and since $n=1$ and $n=3$ gave perfect squares, further values are unlikely to yield perfect squares. 4. **Conclusion:** The only pairs $(n,m)$ are $(1,1)$ and $(3,3)$. **Final answer:** There are 2 such pairs. $$\boxed{2}$$