I posted this answer on BitcoinTalk but will copy it here for those who don't want to chase links. The following is a direct paste from that link:
If you can, installing sage and using that rather than Python will make your life much easier. There is an example notebook on bitcoin.ninja which does some ECDSA stuff on Bitcoin's curve.
To directly answer your question, we can get x from y in basically the same way you got y from x. To see this, let's walk through why your method works.
We have from our curve equation that Y = y^2 = x^3 + 7. You can compute Y easily from x, then you're solving Y = y^2 for y. By Fermat's Little Theorem we can write 1 = y^(p - 1) = Y^(p - 1)/2. Write Q = (p - 1)/2; then we have Y^Q = 1, so Y^(Q + 1) = Y = y^2, so Y^((Q + 1)/2) = y. As it turns out, (Q + 1)/2 = (p + 1)/4, which is why you were able to solve for y by using an exponent of (p + 1)/4. Notice that this depends crucially on p being 3 mod 4; otherwise (p + 1)/4 would not be an integer and we wouldn't be able to compute this. Fortunately our choice of p satisfies this.
OK! So let's do the analogous thing for x. Let's write X = x^3 = y^2 - 7. X can be computed from y easily, so we need to solve X = x^3. Write Q = (p - 1)/3; then X^Q = x^(p - 1) = 1, so X^(Q + 1) = Q = x^3, so X^((Q + 1)/3) = x. As it turns out, (Q + 1)/3 = (p + 2)/9. This time we depend crucially on p being 7 mod 9, in order that this is an integer. Luckily it is! So there you go.
TL;DR use (p + 2)/9 in place of (p + 1)/4.
Oh, and to get the other two cube roots you multiply by a nontrivial cube root of 1. (Similar to you multiplying by -1 in your original code.) One such cube root is 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee.
Here is Python code analogous to yours. It takes one of your output y values and returns the input x value as x2.
y = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
## Field parameters
# Field modulus
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
# Cube root of 1
beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
## Actual code
xcubed = (y*y - 7) % p
print "xcubed = 0x%x" % xcubed
x = pow(xcubed, (p + 2) / 9, p)
print "x1 = 0x%x" % x
print "x2 = 0x%x" % (x * beta % p)
print "x3 = 0x%x" % (x * beta * beta % p)
Its output is
xcubed = 0x4866d6a5ab41ab2c6bcc57ccd3735da5f16f80a548e5e20a44e4e9b8118c26eb
x1 = 0xc994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
x2 = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
x3 = 0xbcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb