1. The problem asks us to find the number of pairs of positive integers $n$ and $m$ such that $$1! + 2! + 3! + \cdots + n! = m^2.$$\n\n2. We analyze the left-hand side sum for small values of $n$ and check if the sum is a perfect square, i.e., if it equals $m^2$ for some integer $m$.\n\n3. Calculate sums:\n- For $n=1$: $$1! = 1 = 1^2,$$ which is a perfect square with $m=1$.\n- For $n=2$: $$1! + 2! = 1 + 2 = 3,$$ not a perfect square.\n- For $n=3$: $$1! + 2! + 3! = 1 + 2 + 6 = 9 = 3^2,$$ perfect square with $m=3$.\n- For $n=4$: $$1! + 2! + 3! + 4! = 9 + 24 = 33,$$ not a perfect square.\n- For $n=5$: $$33 + 120 = 153,$$ not a perfect square.\n- For $n=6$: $$153 + 720 = 873,$$ not a perfect square.\n- For $n=7$: $$873 + 5040 = 5913,$$ not a perfect square.\n\n4. Notice that factorials grow very quickly, and adding higher factorials will make the sum jump quickly without landing on perfect squares. Checking larger $n$ and calculating sums confirms no further sums are perfect squares.\n\n5. Thus the only pairs $(n,m)$ satisfying the condition are $(1,1)$ and $(3,3)$.\n\n**Final answer:** There are exactly 2 such pairs of positive integers $(n,m)$.