Let us assume that the sequence $(x_n)$ has a limit. We shall derive a contradiction (which proves that the limit does not exist).
First, the limit cannot be nonzero or $\pm\infty$, because then $|m+1-nx_n|\to\infty$ by the triangle inequality, hence $x_{n+1}\to 0$. So the limit is zero. Now let us write $r:=m+1$. Then the recursion gives that $$rx_{n+1}-nx_nx_{n+1}=1.$$ Here $rx_{n+1}=o(1)$, hence $nx_nx_{n+1}=-1+o(1)$. In particular, for large $n$, the sign of $x_n$ is alternating (between positive and negative). Changing $n$ to $n-1$ in the above display, we get $$rx_n-(n-1)x_{n-1}x_n=1,$$ and then taking the difference of the two equations, $$r(x_{n+1}-x_n)+(n-1)x_{n-1}x_n-nx_nx_{n+1}=0.$$ That is, $$nx_n(x_{n-1}-x_{n+1})=x_{n-1}x_n+r(x_n-x_{n+1}).$$ Here $x_n$ is nonzero for large $n$, hence $$x_{n-1}-x_{n+1}=\frac{x_{n-1}}{n}+\frac{r}{n}-\frac{rx_{n+1}}{nx_n}=\frac{r+x_{n-1}}{n}-\frac{rnx_nx_{n+1}}{(nx_n)^2}.$$ Using that $x_{n-1}=o(1)$ and $nx_nx_{n+1}=-1+o(1)$, we conclude that $$x_{n-1}-x_{n+1}=\frac{r+o(1)}{n}+\frac{r+o(1)}{(nx_n)^2}.$$ For large $n$, the fractions on the right-hand side are positive, whence $x_{n-1}>x_{n+1}$. Restricting $n$ to odd numbers and even numbers, respectively, we see that both $(x_{2k-1})$ and $(x_{2k})$ are decreasing after omitting finitely many terms. However, this is a contradiction, because one of these sequences is negative after omitting finitely many terms, hence it cannot tend to zero.
Added. Here is an alternative way to derive a contradiction. For large $n$, the last display implies that $$x_{n-1}-x_{n+1}>\frac{r}{2n}.$$ Replacing $n$ by $n+2$, $n+4$, etc., then and adding the resulting inequalities, we get for any integer $l\geq 0$ that $$x_{n-1}-x_{n+1+2l}>\frac{r}{2n}+\frac{r}{2n+2}+\dots+\frac{r}{2n+2l}.$$ This is a contradiction, because under $l\to\infty$, the right-hand side tends to infinity, while the left-hand side tends to $x_{n-1}$.