If we know the GCD and LCM of two integers, can we determine the possible values of these two integers?

If you scale the problem by dividing through by $\rm\:gcd(a,b)\:$ then you are asking how to determine coprime factors of a product. This is equivalent to factoring integers.

Your original question, in the special case $\rm\:gcd(a,b) = lcm(a,b),\:$ is much easier:

Hint $\:$ In any domain $\rm\:gcd(a,b)\ |\ a,b\ |\ lcm(a,b)\ $ so $\rm\:lcm(a,b)\ |\ gcd(a,b)\ \Rightarrow $ all four are associate. Hence all four are equal if they are positive integers. If they are polynomials $\ne 0$ over a field then they are equal up to nonzero constant multiples, i.e. up to unit factors.

If $a \neq b$ then $\gcd(a,b) \leq \min\{a,b\} < \max\{a,b\} \leq \mathrm{lcm}(a,b) = \gcd(a,b)$, contradiction...

Edit: This answer was posted when the question was finding $a,b$ when $\gcd(a,b) = \mathrm{lcm}(a,b)$. For a good answer to the new question, see Bill's answer.

There are $2^m$ possible answers, where $m$ is the number of distinct primes dividing the ratio of LCM to GCD. If $$ \eqalign{ \text{gcd}(x,y) &= \prod p_i^{a_i} \\ \text{lcm}(x,y) &= \prod p_i^{b_i} } $$ (with $a_i\le b_i$) and $$ \eqalign{ x &= \prod p_i^{c_i} \\ y &= \prod p_i^{d_i} } $$ then each $c_i$ is either $a_i$ or $b_i$, and each $d_i$ must be the other one. This is because $a_i=\min(c_i,d_i)$ and $b_i=\max(c_i,d_i)$. To invert this process, we have more than one solution whenever $a_i\ne b_i$. This gives two choices for each of the $m$ values of $i$ where $a_i<b_i$, or $2^m$ total possibilities.