Lets call the first of the consecutive integers N, and call the square S^2.  And the number of digits D (D=3 for this puzzle).

Then (10^D + 1) * N + 1 = S^2.  

Let P be some prime factor of 10^D + 1.  

Now take the first equation mod P:  1 = S^2 mod P

Rearrange a little bit: (S-1) * (S+1) = 0 mod P

Now because P is prime the only solutions (mod P) are the obvious solutions: S = 1 mod P and S = P-1 mod P.

(I am assuming 10^D+1 is squarefree)

From this we can take each prime factor of 10^D + 1 and create a system of residues to solve, which can be done with the Chinese remainder theorem.  

The system with all remainders equal to 1 is discarded since all that does is make S=1.  

Similarly, the system with all remainders 1 less than the respective primes (p-1 for each prime p) is discarded since all that does is make S=10^D, which is too large.

So for this puzzle D=3: 10^D + 1 = 1001 = 7*11*13

Then six candidates for S emerge:

S=6 mod 7; S=1 mod 11; S=1 mod 13 yields S=573: S^2=328329

S=1 mod 7; S=10 mod 11; S=1 mod 13 yields S=274: S^2=75076 (leading zero)

S=6 mod 7; S=10 mod 11; S=1 mod 13 yields S=846: S^2=715716

S=1 mod 7; S=1 mod 11; S=12 mod 13 yields S=155: S^2=24025 (leading zero)

S=6 mod 7; S=1 mod 11; S=12 mod 13 yields S=727: S^2=528529

S=1 mod 7; S=10 mod 11; S=12 mod 13 yields S=428: S^2=183184

If we go further to D=4: 10^4+1 = 10001 = 73*137

Two candidates:

S=1 mod 73; S=136 mod 137 yields S=2191: S^2=4800481 (leading zero)

S=72 mod 73; S=1 mod 137 yields S=7810: S^2=60996100

Lets go higher than Charlie's program went and try D=8: 10^8+1 = 100000001 = 17*5882353

S=1 mod 17; S=1 mod 5882353 yields S=47058823: S^2=2214532822145329

S=16 mod 17; S=5882352 mod 5882353 yields S=52941178: S^2=2802768328027684