In the normal setting $n=pq$ is public knowledge and $\varphi(n)$ is hidden, for a start.

I will assume $$ed\equiv 1 \pmod {\varphi(n)}\quad(1).$$ Since

$$\varphi{(n)} = (p - 1)(q - 1) = pq - p - q + 1 = (n + 1) - (p + q)$$

Also, $n = pq$ and some manipulation gives

$$n = p \left ( n + 1 - \varphi{(n)} - p \right ) = -p^2 + (n + 1 - \varphi{(n)})p$$ and then $$p^2 - (n + 1 - \varphi{(n)})p + n = 0$$

which can be solved by the quadratic formula for $p.$ In conclusion, knowledge of $\varphi{(n)}$ allows one to factor $n$ in constant time.

But we don’t know $n$ and we only know $ed-1=k\varphi(n)$ for some positive integer $k$ from (1).

We can look for small divisors of $ed-1,$ since $k$ may have small divisors in an attempt to find $\varphi(n).$ This may give us a few small divisors but it may not be enough to determine $ed-1.$

A related question might be, how many $n$ would we expect to be compatible with the given $(e,d)$ pair. This is equivalent to scanning through multiples of $z=ed-1$ looking for semiprimes.

There may be many semiprimes $p_iq_i$ among those $\{kz:1\leq \}$ but we also require that $p_i,q_i$ be not much smaller than $\sqrt{n}$ for a well chosen RSA modulus $n$. If we also know the bitlength of $n$ there may not be that many candidates. From Green and Tao’s work on primes in arithmetic progressions we know that gaps between primes in the same AP are huge, but these are semiprimes.

In the normal setting $n=pq$ is public knowledge and $\varphi(n)$ is hidden, for a start.

I will assume $$ed\equiv 1 \pmod {\varphi(n)}\quad(1).$$ Since

$$\varphi{(n)} = (p - 1)(q - 1) = pq - p - q + 1 = (n + 1) - (p + q)$$

Also, $n = pq$ and some manipulation gives

$$n = p \left ( n + 1 - \varphi{(n)} - p \right ) = -p^2 + (n + 1 - \varphi{(n)})p$$ and then $$p^2 - (n + 1 - \varphi{(n)})p + n = 0$$

which can be solved by the quadratic formula for $p.$ In conclusion, knowledge of $\varphi{(n)}$ allows one to factor $n$ in constant time.

But we don’t know $n$ and we only know $ed-1=k\varphi(n)$ for some positive integer $k$ from (1).

We can look for small divisors of $ed-1,$ since $k$ may have small divisors in an attempt to find $\varphi(n).$ This may give us a few small divisors but it may not be enough to determine $\varphi(n).$

However [see comments] this actually leaves only a few possibilities for $k$ and thus for we can quickly determine $\varphi(n)$.