This problem can be recast in terms of the famous problem of the number of ways to represent a positive integer as a sum of squares. With this perspective, we can see that the following more general statement is true for any $p > 1$ (so that each of the infinite series actually converges):
$$\sum_{n=1}^{\infty} \frac{1}{(n^2)^p} + \sum_{m,n = 1}^{\infty} \frac{1}{(m^2+n^2)^p} = \left(\sum_{n=1}^{\infty} \frac{1}{n^p}\right) \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^p}\right).$$
The left-hand side is
$$\sum_{s=1}^{\infty} \frac{n_2(s)}{s^p},$$ where $n_2(s)$ is the number of ways of representing $s$ as the sum of one or of two squares of positive integers, in which order is distinguished (i.e., $1^2 + 2^2$ is counted separately from $2^2+1^2$).
It is known that $n_2(s) = d_1(s) - d_3(s)$ (see eq. 24 in the site linked above), where $d_k(s)$ is the number of divisors of $s$ congruent to $k \bmod 4$.
The first sum on the right side of the equation has (the $p$th powers of) all positive integers as denominators and the second sum on the right has (the $p$th powers of) all odd numbers as denominators. After multiplying those sums together, then, $1/s^p$ (ignoring signs) appears on the right as many times as there are odd divisors of $s$. Each odd divisor of $s$ congruent to $1 \bmod 4$ contributes a $+1/s^p$, and each odd divisor of $s$ congruent to congruent to $3 \bmod 4$ contributes a $-1/s^p$. Thus the coefficient of $1/s^p$ on the right side is exactly $d_1(s) - d_3(s)$. Therefore, the right-hand side is also
$$\sum_{s=1}^{\infty} \frac{n_2(s)}{s^p}.$$