While tinkering with Mathematica I found that it could evaluate sums of the form $$\sum\limits_{k = 1}^{\infty} \frac{H_k H_{k+m}}{k^2}$$ for some nonnegative integer $m$. As it turns out (or the evidence is overwhelmingly in favor of it), this sum is always a linear combination of a constant term and the first three $\zeta$-values, i.e. $\zeta(2)$ through $\zeta(4)$. The coefficients of the $\zeta$-values were not too hard to figure out with the help of the OEIS and a bit of intuition. We have $$\sum\limits_{k = 1}^{\infty} \frac{H_k H_{k+m}}{k^2} = -C -\left(\sum\limits_{k = 1}^{m}\frac{1}{k^2}\right)\zeta(2) + 2\left(\sum\limits_{k = 1}^{m}\frac{1}{k}\right)\zeta(3) + \frac{17}{4}\zeta(4)$$ with just the non-zeta term $C$ being undetermined. For $m = 1,\ldots,9$ this constant takes on the values $$C = 0,\frac{1}{4},\frac{4}{9},\frac{341}{576},\frac{679}{960},\frac{25921}{32400},\frac{19879}{22680},\frac{95594629}{101606400},\frac{182134073}{182891520}$$ which... isn't all that insightful, to say the least. Visually, this looks like
Thanks to the comment by ho boon suan (and verification by Dr. Wolfgang Hintze) I was able to verify that indeed $$C = \frac{1}{2} \sum\limits_{j = 1}^{m} \frac{H_{j-1}^2+H_{j-1}^{(2)}}{j^2}$$
and thus
$$\sum\limits_{k = 1}^{\infty} \frac{H_k H_{k+m}}{k^2} = -\sum\limits_{j = 1}^{m} \underbrace{\frac{H_{j-1}^2+H_{j-1}^{(2)}}{2j^2}}_{=\frac{1}{j^2(j-1)!} \begin{bmatrix}j \\ 3\end{bmatrix}} -H_m^{(2)}\zeta(2) + 2H_m\zeta(3) + \frac{17}{4}\zeta(4)$$
As a side product of an auxiliary calculation, we also have the identity
$$\sum\limits_{k = 1}^{\infty} \frac{H_k}{k^2(k+j)} = \frac{H_{j-1}^2+H_{j-1}^{(2)}}{j^2} - \frac{1}{j^2} \zeta(2) + \frac{2}{j}\zeta(3)$$
for arbitrary $j \in \mathbb{N}$.
EDIT 13.02.23
I am grateful to Ali Shadhar for a very helpful conversation which allowed me to improve part of the discussion section.
EDIT 11.02.23
I am grateful to Mariusz Iwaniuk who pointed out a sign mistake in $(3)$, which I have now corrected. The original author of the OP, TheOutZ, already had the term correct.
Original post
The sum
$$s(m)=\sum _{k=1}^{\infty } \frac{H_k H_{k+m}}{k^2}\tag{1}$$
can be simplified in terms of finite sums writing
$$H_{k+m}=\sum _{i=1}^m \frac{1}{i+k}+H_k\tag{2}$$
then expanding the product and doing the sums with the result
$$s(m) =-\zeta (2) H_m^{(2)}+2 \zeta (3) H_m+\frac{17 \zeta (4)}{4} -\frac{1}{2} \sum _{s=1}^m \frac{1}{s^2}( H_{s-1}{}^2+H_{s-1}^{(2)})\tag{3}$$
so that your "constant" is the rational number given by
$$C_m = \frac{1}{2} \sum _{s=1}^m \frac{1}{s^2}( H_{s-1}{}^2+H_{s-1}^{(2)})\tag{4}$$
Mathematica finds $C_\infty = \frac{7\pi^4}{360} \simeq 1.89407$
Discussion
Some possibly useful relations are.
The combinations of $H$ in the summands in $C$ can be written as
$$c(n)=\left(H_{n}\right){}^2+H_{n}^{(2)} = 2 \sum_{i=1}^{n}\frac{1}{i^2} + 2\sum_{1\le i\lt j \le n}\frac{1}{i j}$$
its generating function
$$\sum_{n=1}^ {\infty}(\left(H_{n}\right){}^2+H_{n}^{(2)}) x^n =\frac{2 \text{Li}_2(x)}{1-x}+\frac{\log ^2(1-x)}{1-x} \tag{5}$$
Here we have combined
$$\sum_{n=1}^ {\infty}(H_{n}^{(2)}) x^n =\frac{ \text{Li}_2(x)}{1-x} $$
and
$$\sum_{n=1}^ {\infty}(H_n^2 - H_{n}^{(2)}) x^n=\frac{\log ^2(1-x)}{1-x} $$
(see Ali's book, 2.1.5)
Generalization to non-integer $m$
As a first test I tried to calculate $s(\frac {1}{2}) \simeq 5.01804$.
With the help of Ali and drawing from his book we could find the closed expression
$$s(\frac{1}{2}) = 8 \text{Li}_4\left(\frac{1}{2}\right)+4 \zeta (3)+\zeta (3) \log (8)-\frac{\pi ^4}{90}\\ +\frac{\log ^4(2)}{3}-\frac{1}{3} \pi ^2 \log ^2(2)-8 \log ^2(2)\tag{6}$$
Indeed, writing e.g. (https://math.stackexchange.com/a/3035646/198592) $$H_{n+\frac{1}{2}}=2 H_{2 n+1}-H_n-\log (4)\tag{7}$$
we encounter the in $s(\frac{1}{2})$ the sums over $ H_n H_{2n}/n^2, H_n^2/n^2, H_n/(n^2(2n+1)), H_n/n^2$ which are all solved in Ali's book.
For a general small $m=\alpha$ we can form a power series expansion of $s(\alpha)$ based on
$$\frac{\partial H_{n+\alpha }}{\partial \alpha }\text{/.}\, \alpha \to 0 = \frac{\pi ^2}{6}-H_n^{(2)}\tag{8}$$
leads to the sum
$$s_2 = \sum _{n=1}^{\infty } \frac{H_n H_n^{(2)}}{n^2}=\zeta(5)+\zeta(3)\zeta(2)\tag{9}$$
which was calculated in Ali's book, formula $(4.107)$.
Summarizing, the case of non-integer $m$ promises to lead to interesting problems in terms of sums (or the equivalent integrals).
The case of rational $\alpha$ can perhaps be solved.