I don't know if there is a closed form for k, but a numerical answer accurate to 6 or more decimal places will be fine.
This seems numerically unstable, so computationally challenging.
Sqdancefan
In order for there to be exactly two real solutions, one of the intersection points between y=x^2 and y=k^x will be where the two curves are tangent to each other. That is, both the y values and the y' values will be equal at that point. This means we are solving the simultaneous equations ...
.. x^2 = k^x
.. 2x = (k^x)*ln(k)
Using the former in the latter, we have
.. 2x = x^2*ln(k)
.. (2/x) = ln(k)
.. k = e^(2/x)
Using this in the first equation, we have
.. x^2 = (e^(2/x))^x
.. x^2 = e^2
.. x = e
So, the value of k is e^(2/e), and the tangent point is (e, e^2).
.. k ≈ 2.0870652286345329598
_____
The other point of intersection is x=-e*ProductLog[1/e] ≈ -0.75694510645758
JB
This is not an answer, just a test, because one answer that was here earlier has now disappeared. I may contact the answerer to see if he/she can enter it again, because I came here just now to award him/her best answer .... but it is gone! A Y!A bug, I guess.
Here is the exact answer that J wrote that has now disappeared. I found the answer quite interesting because I had no idea that the value of k could be given exactly in closed form. When I contacted him he said it had not disappeared for him. Hmmm. Here is J's answer:
Nice problem. After at least an hour working on this I found the exact answer, namely k = e^(2/e) ≈ 2.087065228634533. The solutions for that value of k are e and a number whose approximate value is -0.756945106457583664584. The double root is x=e.
Sorry, but the derivation details are too much for me to type in here.