Note that $(n + m - 1)|(n + m - 1)^2 + 2(n + m - 1) = (n + m - 1)^2 + 2mn$. By combining it with given condition we get $(n + m - 1)|2mn$. Since none of $(n + m - 1)$ and $(n + m + 1)$ equals to $2$, greatest common divisor (GCD) of these numbers is not greater than $2$.
If $GCD = 2$ then $\frac{1}{2}(n + m - 1)(n + m + 1)|2mn$. Also if $GCD = 1$ then $(n + m - 1)(n + m + 1)|2mn$. From here we deduce that $2mn \neq 0$, $\frac{1}{2}(n + m - 1)(n + m + 1) \le 2mn$ and $(n + m)^2 - 1 \le 4mn \Rightarrow (n + m)^2 \le 1$ $(*)$
Since $n, m$ symmetrical, we may assume $n \ge m$. Considering the $(*)$, we get either $n = m$ or $n = m + 1$

a. Let $n = m$. By the given condition $2m + 1|2m^2$ and $(2m + 1, m) = 1$, $(2m + 1, 2) = 1$. It is obvious that there are no such numbers.

b. Let $n = m + 1$. By the given conditions
$$\begin{cases} (n + m + 1)|2mn \\ (n + m - 1)|n^2 + m^2 - 1 \end{cases} \Leftrightarrow \begin{cases} 2(m + 1)|2mn(m + 1) \\ 2m|2mn(m + 1) \end{cases}$$
and we conclude that any natural $m$ satisfies the conditions.

Thus $(n, m)$ is pair of successive natural numbers.