Fourier or Taylor?

Note that $\begin{aligned} \frac{m! n!}{(m+n+2)!} & = \; \frac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+3)} \; = \; \frac{1}{m+1} \frac{\Gamma(m+2)\Gamma(n+1)}{\Gamma(m+n+3)} \;= \; \frac{1}{m+1}B(m+2,n+1) \\ & = \; \frac{1}{m+1}\int_0^1 x^{m+1}(1-x)^n\,dx \end{aligned}$ for any $m,n \ge 0$ , and hence $\sum_{n \ge 0} \frac{m! n!}{(m+n+2)!} \; = \; \frac{1}{m+1}\int_0^1 x^{m+1} \frac{1}{1 - (1-x)}\,dx \; = \; \frac{1}{m+1} \int_0^1 x^m\,dx \; = \; \frac{1}{(m+1)^2}$ making the double sum $\sum_{m,n \ge 0} \frac{m! n!}{(m+n+2)!} \; = \; \sum_{m \ge 0} \frac{1}{(m+1)^2} \; = \; \sum_{k \ge 1} k^{-2} \; = \; \tfrac16\pi^2$

We have that $\begin{aligned} \frac{n!}{(m + n + 1)!} - \frac{(n + 1)!}{(m + n + 2)!} &= \frac{n!}{(m + n + 2)!} [(m + n + 2) - (n + 1)] \\ &= \frac{n! (m + 1)}{(m + n + 2)!}, \end{aligned}$ so $\frac{n!}{(m + n + 2)!} = \frac{1}{m + 1} \left( \frac{n!}{(m + n + 1)!} - \frac{(n + 1)!}{(m + n + 2)!} \right).$

Hence, following sum telescopes: $\sum_{n = 0}^\infty \frac{n!}{(m + n + 2)!} = \sum_{n = 0}^\infty \frac{1}{m + 1} \left( \frac{n!}{(m + n + 1)!} - \frac{(n + 1)!}{(m + n + 2)!} \right) = \frac{1}{(m + 1) (m + 1)!}.$ Then $\begin{aligned} \sum_{m = 0}^\infty \sum_{n = 0}^\infty \frac{m! n!}{(m + n + 2)!} &= \sum_{m = 0}^\infty m! \sum_{n = 0}^\infty \frac{n!}{(m + n + 2)!} \\ &= \sum_{m = 0}^\infty m! \cdot \frac{1}{(m + 1) (m + 1)!} \\ &= \sum_{m = 0}^\infty \frac{1}{(m + 1)^2} \\ &= \sum_{k = 1}^\infty \frac{1}{k^2}. \end{aligned}$