Find all natural numbers $k$ for which there exist natural numbers $n,m$ such that $m(m+k) = n(n+1)$.
Rearranging the given equation gives $k = \dfrac{n(n+1)}{m}-m$. Thus, $m \mid n(n+1)$ and so we have cases.
Case 1: $n = am$ for some $a \in \mathbb{Z}^+$ In this case we have $k = a(am+1)-m$, which is positive for all $a,m$.
Case 2: $n+1= am$ for some $a \in \mathbb{Z}^+$ In this case we have $k = a(am-1)-m$ and we need $k$ to be positive and so $a(am-1)-m \geq 1$, or $am \geq m+1$. This is true if and only if $a \geq 2$.
I didn't see how to solve the other case where both are not multiples of $m$.
The equation
$$m(m+k) = n(n+1)$$
has positive integer solutions $m,n$ for all positive integers $k$ except $k = 2,3$.
Proof:
If $k = 1$, just let $m = n$, where $n$ is an arbitrary positive integer.
If $k$ is even, $k > 2$, let
$$m = (k(k-2))/4$$
$$n = ((k+2)(k-2))/4$$
and if $k$ is odd, $k > 3$, let
$$m = ((k-1)(k-3))/8$$
$$n = ((k+3)(k-3))/8$$
In both cases, the equation $m(m+k) = n(n+1)$ is identically satisfied.
Next, suppose $k = 2$. Then
$$m(m + 2) = n(n + 1)$$
implies $m < n$ and $m > n-1$, contradiction.
Finally, suppose $k = 3$. Then
$$m(m + 3) = n(n + 1)$$
implies $m < n$ and $m > n -2$, hence $m = n -1$. But then
$$m(m + 3) = n(n+1) \implies (n-1)(n+2) = n(n+1) \implies n = \tfrac{1}{2}$$
contradiction.
This completes the proof.