Claim 1. There exists infinitely many $m$ such that set $\{ m+1, \ldots, m+n^2 \}$ contains at most $n$ Jane's integers. The below proof gives stronger claim that set $\{ m+1, \ldots, m+n^2 \}$ can contain no Jane's integers. Proof. If there is a Jane's integer $x^3+y^2$ with $x \ne t$ in $(t^3+1,(t+1)^3)$ then $x<t$. Let $x=t-i$ then from $x^3+y^2>t^3$ we find $y^2>3t^2i-3ti^2+i^3$ or $4y^2>i \left[ 3(2t-i)^2+i^2 \right] \ge 3it^2$ or $y>\frac{t\sqrt{3i}}{2}$. We follow \begin{align*} x^3+\left( y+ \frac{t\sqrt{3}}{\sqrt i} \right)^2 & =x^3+y^2+ \frac{2yt\sqrt{3}}{\sqrt{i}}+ \frac{3t^2}{i}, \\ & > t^3+1+ 3t^2+\frac{3t^2}{i} \ge (t+1)^3. \end{align*}Therefore, there are at most $\left \lfloor \frac{t\sqrt{3}}{\sqrt i} \right \rfloor$ Jane's integers $(t-i)^3+y^2$ in $(t^3+1,(t+1)^3)$ for $1 \le i \le t-1$. For $i=0$ then since $t^3+(1+\sqrt{3}t)^2>(t+1)^3$ so there are at most $\left \lfloor \sqrt{3}t \right \rfloor$ Jane's integers $t^3+y^2$ in $(t^3+1,(t+1)^3)$. Thus, number of Jane's integers in $(t^3+1,(t+1)^3)$ is at most $$\left \lfloor \sqrt{3}t \right \rfloor + \sum_{i=1}^{t-1} \left \lfloor t\sqrt{\frac 3i} \right \rfloor \le \sqrt{3}t \left(1+\sum_{i=1}^{t-1}\frac{1}{\sqrt{i}} \right)< \sqrt{3}t(1+2\sqrt{t})<4t^{3/2}.$$(The proof for the bound $\sum_{i=1}^{t-1}\frac{1}{\sqrt i}<2\sqrt{t}$ can be found here). But since $(t+1)^3-t^3=3t^2+3t>3t^2$ and there are at most $4t^{3/2}$ Jane's integers in $(t^3+1,(t+1)^3)$ so there exists two consecutive Jane's integers $a,b$ so that their difference is at least $\frac{3t^2}{4t^{3/2}}=\frac 34 t^{1/2}$. Pick larger enough $t$ and choose $n^2$ consecutive integers in $(a,b)$ and we are done. $\square$ Claim 2. There exists infinitely many $m$ such that set $\{ m+1, \ldots, m+n^2 \}$ contains at least $n+1$ Jane's integers. Proof. Pick $m=t^3$ then we can see that the set $\{ t^3+1, \ldots, t+n^2 \}$ already has $n$ Jane's integers $t^3+1,t^3+2^2, \ldots, t^3+n^2$. It suffices to choose $t$ so $t^3+2$ is also a Jane's integer, or $t^3+2=(t-1)^3+y^2$ for some $y \ge 1$. This is equivalent to $3(t^2-t+1)=y^2$. Let $y=3z$ then $(2t-1)^2+3=3(2z)^2$. Thus, $3 \mid 2t-1$ so $2t-1=3k$ and we have $(2z)^2-3k^2=1$. This is Pell's equation $x^2-3y^2=1$ with fundamental solution $(x,y)=(2,1)$ and has infinitely many solutions $(u_n,v_n)_{n \ge 1}$: $$u_{n+1}=2u_n+3v_n, v_{n+1}=u_n+2v_n.$$From this, we find that there exists infinitely many solutions $(x,y)$ for Pell's equation so $2 \mid x$, which follows there exists infinitely many $t,y$ so $t^3+2=(t-1)^3+y^2$ or $t^3+2$ is Jane's integer. $\square$ Back to the problem, let $S(m)$ be number of Jane's integers in set $\{ m+1, \ldots, m+n^2 \}$. We can easily see that $S(m+1)-S_m \in \{ \pm 1,0 \}$. Now, from two above claims, say we have $m_1,m_2$ so $S(m_1) \le n$ and $S(m_2) \ge n+1$ then there must exists $m_1<M\le m_2$ so $S(M)=n+1$. Since there exists infinitely such $m_1,m_2$ so the exists infinitely $M$. We are done.

For our lack of affinity with Jane, call such numbers good instead. For each $m \geqslant 1$ let $f(m)$ denote the number of good numbers in the set $\{m+1, \dots, m+n^2\}$. Note that $f(m+1)-f(m) \in \{-1, 0, 1\}$ so $f$ is "discretely continuous", thus it suffices to show that both $f(m)<n+1$ and $f(m) \geqslant n+1$ hold infinitely often to get the desired result. To see $f(m) \geqslant n+1$ hold infinitely often, consider $m=r^3$ for some integer $r \geqslant 1$. Notice that each of the numbers $m+j^2$ for $1 \leqslant j \leqslant n$ is good. Also, let $r$ satisfy the Pell-ish equation, $r^2-r+1=3s^2$. Then, $$r^3+2=(r-1)^3+(3s)^2,$$so we get another good number, making $f(m) \geqslant n+1$ here. To see $f(m)=0$ is possible infinitely often, take the interval $[1, x]$ and observe that $$a^3+b^2=c \leqslant x \Longrightarrow (a, b) \in [1, x^{\frac{1}{3}}] \times [1, x^{\frac{1}{2}}],$$so the density of good numbers is at best $\lim_{x \rightarrow \infty} \frac{x^{\frac{5}{6}}}{x}=0.$ However, if $f(m) \geqslant 1$ for all but finitely many $m$, then this density is $\geqslant \frac{1}{n^2}>0,$ contradiction!