I found a proof relying on some results from number theory, which hopefully fits your requirements
Claim:
$$
\left(\sum_{n \geq 1}\frac{(-1)^n}{2n+1}\right)^2=\frac{3}{8}\sum_{n\geq1}\frac{1}{n^2}
$$
We define
- $\chi_l(n)$ as the l'th Dirichlet character $\text{mod}\,4$
- $L(\chi,s)\equiv\sum_{n\geq1}\tfrac{\chi(n)}{n^s}$ is a Dirichlet-$L$ sum
- $\zeta(s,q)\equiv\sum_{n\geq0}\tfrac{1}{(n+q)^s}$ is a Hurwitz zeta function ($q=1$ gives Riemanns $\zeta(s)$)
Now we recoginze that the orignal problem can be reformulated as follows
$$
L^2(1,\chi_1)=\frac{1}{2}L(2,\chi_0) \quad(*)
$$
Proof:
By virtue of the identity (this holds for general characters $\text{mod}\,\,a $)$ L(\chi,s)=\sum_{b\leq a}\chi(b)\zeta(s,\frac ba)$ and the explict values of the character table we get
for the left hand side of $(*)$
$$
\frac{1}{16}\left(\zeta(1,\tfrac{1}{4})-\zeta(1,\tfrac{3}{4})\right)^2
$$
where the limit $s\rightarrow 1$ is implicitly taken. By the Stieltjes expansion of the Hurwitz Zeta function this equal to
$$
\frac{1}{16}(\psi_0(1/4)-\psi_0(3/4))^2=\frac{\pi^2}{16}\cot^2(\frac{\pi}{4}) \quad(**)
$$
where the equality is a consequence of the reflection formula for the Polygamma function $\psi_n(z)$.
On the other hand, for the rhs of $(*)$ we can write by the nearly the same token (here no limiting procedure is necessary)
$$
\frac{1}{2\cdot16}\left(\psi_1(1/4)+\psi_1(3/4)\right)=\frac{\pi^2}{2\cdot16}\frac{1}{\sin^2(\frac{\pi}4)}\quad(***)
$$
Now since $\cos^2(\frac{\pi}{4})=\frac{1}{2}$, $(**)=(***)$ and therefore $(*)$ is proven
Since i'm a theoretical physicist my knowledge of number theory is close to zero. I guess a more experienced person could conclude here much faster using some general theorems of $L$-functions or Dirichlet convolution.