For the case $m=k=2$, in which we seek the number $N(n)$ of pairs $(x,y) \in [1,n]^2$ for which $xy$ is a square, we give an elementary estimate $$ N(n) = Cn \log n + An + O(n^{2/3}), $$ where $C = 1/\zeta(2) = 6/\pi^2$ and $$ A = \frac{3\gamma-1}{\zeta(2)} - \frac{2\zeta'(2)}{\zeta(2)^2} - 1 = 0.1377775\ldots \, . $$ This agrees with the analytic calculation of Asymptotiac K (and is corroborated by numerical computation up to $n = 2^{30}$), and improves the error term (for which the contour-integral error term, equivalent to $n^{21/22+o(1)}$ rather than $n^{2/3}$).

Recall that $xy$ is a square iff $(x,y) = (ma^2,mb^2)$ for some positive integers $m,a,b$, and the representation can be made unique by requiring either that $a,b$ be coprime or that $m$ be squarefree, which is the source of the factor $1/\zeta(2)$. Let $M(n)$, then, be the number of triples $(m,a,b)$ of positive integers such that $ma^2\leq n$ and $mb^2\leq n$, without the additional coprimality or squarefree condition. Then Möbius inversion (applied with either condition $\gcd(a,b)=1$ or $\mu(m)^2 = 1$ yields $$ N(n) = \sum_{d=1}^{\lfloor\sqrt{n}\rfloor} \mu(d) M(\lfloor n/d^2 \rfloor). $$ We show:

Proposition. $M(n) = n \log n + Bn + O(n^{2/3})$, where $B = 3\gamma - 1 - \zeta(2) = -0.913287\ldots$ (and again $\gamma$ is Euler's constant $0.5772156649\ldots$).

Proof: Let $R = \lfloor n^{1/3} \rfloor$. If $m\cdot\max(a,b)^2 \leq n$ then either $m \leq R$ or $\max(a,b) \leq R$. Given $m$, the number of $(a,b)$ pairs is $\lfloor \sqrt{n/m} \rfloor^2 = n/m - O(\sqrt{n/m})$; summing this over $m \leq R$ gives $n H_R - O(n^{2/3})$ where $H_R$ is the harmonic sum $\sum_{m=1}^r 1/m$. In the other direction, each $k \leq R$ occurs $2k-1$ times as $\max(a,b)$ for positive integers $a,b$, each of which accounts for $\lfloor n/k^2 \rfloor$ solutions, for a sum of $$ \sum_{k=1}^R (2k-1) \, (n/k^2 - O(1)) = (2 H_R - \zeta(2)) n - O(n^{2/3}) $$ solutions. For the total count we add this to $n H_r - O(n^{2/3})$, and subtract $R^3 = n - O(n^{2/3})$ which is the number of solutions for which $a,b,m$ are all $\leq R$, finding $(3 H_R - \zeta(2) - 1) n + O(n^{2/3})$. The Proposition then follows from $H_R = \log R + \gamma + O(1/R) = \frac13 \log n + \gamma + O(n^{-1/3})$. $\Box$

The estimate for $N(n)$ then follows from $N(n) = \sum_{d=1}^{\lfloor\sqrt{n}\rfloor} \mu(d) M(\lfloor n/d^2 \rfloor)$; the $2\zeta'(2) / \zeta(2)^2$ comes from the second term of $$ \frac{n}{d^2} \log \frac{n}{d^2} = \frac1{d^2} n \log n - 2n \frac{\log d}{d^2} $$ because $\sum_{d=1}^\infty \mu(d) \log d \, / \, d^2$ is the derivative at $s=2$ of $-1/\zeta(s)$.

One can probably use bounds on exponential sums to further reduce the $O(n^{2/3})$ error, both in worst and average case, as is done for the Dirichlet divisor problem and the Gauss circle problem.

For the case $m=k=2$, in which we seek the number $N(n)$ of pairs $(x,y) \in [1,n]^2$ for which $xy$ is a square, we give an elementary estimate $$ N(n) = Cn \log n + An + O(n^{2/3}), $$ where $C = 1/\zeta(2) = 6/\pi^2$ and $$ A = \frac{3\gamma-1}{\zeta(2)} - \frac{2\zeta'(2)}{\zeta(2)^2} - 1 = 0.1377775\ldots \, . $$ This agrees with the analytic calculation of Asymptotiac K (and is corroborated by numerical computation up to $n = 2^{30}$), and improves the error term (the contour-integral analysis gave an error estimate equivalent to $n^{21/22+o(1)}$ rather than $n^{2/3}$).

Recall that $xy$ is a square iff $(x,y) = (ma^2,mb^2)$ for some positive integers $m,a,b$, and the representation can be made unique by requiring either that $a,b$ be coprime or that $m$ be squarefree, which is the source of the factor $1/\zeta(2)$. Let $M(n)$, then, be the number of triples $(m,a,b)$ of positive integers such that $ma^2\leq n$ and $mb^2\leq n$, without the additional coprimality or squarefree condition. Then Möbius inversion (applied with either condition $\gcd(a,b)=1$ or $\mu(m)^2 = 1$ yields $$ N(n) = \sum_{d=1}^{\lfloor\sqrt{n}\rfloor} \mu(d) M(\lfloor n/d^2 \rfloor). $$ We show:

Proposition. $M(n) = n \log n + Bn + O(n^{2/3})$, where $B = 3\gamma - 1 - \zeta(2) = -0.913287\ldots$ (and again $\gamma$ is Euler's constant $0.5772156649\ldots$).

Proof: Let $R = \lfloor n^{1/3} \rfloor$. If $m\cdot\max(a,b)^2 \leq n$ then either $m \leq R$ or $\max(a,b) \leq R$. Given $m$, the number of $(a,b)$ pairs is $\lfloor \sqrt{n/m} \rfloor^2 = n/m - O(\sqrt{n/m})$; summing this over $m \leq R$ gives $n H_R - O(n^{2/3})$ where $H_R$ is the harmonic sum $\sum_{m=1}^r 1/m$. In the other direction, each $k \leq R$ occurs $2k-1$ times as $\max(a,b)$ for positive integers $a,b$, each of which accounts for $\lfloor n/k^2 \rfloor$ solutions, for a sum of $$ \sum_{k=1}^R (2k-1) \, (n/k^2 - O(1)) = (2 H_R - \zeta(2)) n - O(n^{2/3}) $$ solutions. For the total count we add this to $n H_r - O(n^{2/3})$, and subtract $R^3 = n - O(n^{2/3})$ which is the number of solutions for which $a,b,m$ are all $\leq R$, finding $(3 H_R - \zeta(2) - 1) n + O(n^{2/3})$. The Proposition then follows from $H_R = \log R + \gamma + O(1/R) = \frac13 \log n + \gamma + O(n^{-1/3})$. $\Box$

The estimate for $N(n)$ then follows from $N(n) = \sum_{d=1}^{\lfloor\sqrt{n}\rfloor} \mu(d) M(\lfloor n/d^2 \rfloor)$; the $2\zeta'(2) / \zeta(2)^2$ comes from the second term of $$ \frac{n}{d^2} \log \frac{n}{d^2} = \frac1{d^2} n \log n - 2n \frac{\log d}{d^2} $$ because $\sum_{d=1}^\infty \mu(d) \log d \, / \, d^2$ is the derivative at $s=2$ of $-1/\zeta(s)$.

One can probably use bounds on exponential sums to further reduce the $O(n^{2/3})$ error, both in worst and average case, as is done for the Dirichlet divisor problem and the Gauss circle problem.