This sequence $\lfloor \sqrt{2003}\cdot n\rfloor $ contains an infinite number of square numbers

Assume that $\frac{p}{q}$ is a convergent of the continued fraction of $\sqrt{2003}$. We have $$ \left|\sqrt{2003}-\frac{p}{q}\right|\leq \frac{1}{q^2} $$ hence $n=pq$ (or $n=pq\pm 1$) is a good candidate for $\left\lfloor n\sqrt{2003}\right\rfloor$ to be a square, since $$ \left| pq\sqrt{2003}-p^2\right|\leq \frac{p}{q}\approx\sqrt{2003}. $$ You just have to show that among the best rational approximations given by the continued fraction of $\sqrt{2003}$, there are some "very best" approximations for which $$ \left|\sqrt{2003}-\frac{p}{q}\right|\leq \frac{C}{q^2} $$ holds with a very small constant $C$.
That can be done by explicitly writing the continued fraction of $\sqrt{2003}$ $$\sqrt{2003}=[44;\overline{1,3,12,1,1,6,2,1,2,1,3,6,7,1,\color{red}{43},1,7,6,3,1,2,1,2,6,1,1,12,3,1,\color{red}{88}}]$$ and studying what happens at each step, especially when we meet a large element of such continued fraction.

Yes, using the Pell equation is a great idea! In this case, the best Pell equation to use is $x^2-2003y^2 = -2$ (there are no solutions to the $-1$ version of this Pell equation). Given such a solution, note that $$ xy\sqrt{2003} - x^2 = \frac{2x}{y\sqrt{2003} + x} = \frac{2x}{\sqrt{x^2+2} + x}, $$ and therefore $$ 0 < xy\sqrt{2003} - x^2 < 1, $$ showing that $\lfloor xy\sqrt{2003} \rfloor = x^2$.

The first solution to the Pell equation $x^2-2003y^2 = -2$ is $(x,y) = (65912269,1472739)$, and an infinite family of solutions can be generated in the usual way using the fundamental solution $(x_1,y_1) = (4344427204728362,97071569134791)$ to $x^2-2003y^2 = 1$. (And indeed, these two solutions come exactly from the convergents indicated by the red numbers in Jack D'Aurizio's answer!)

Make this CW... the business with $xy \sqrt d$ does not work, as such, for $\sqrt 7.$ However, we seem to have no trouble finding $n$ for which $\lfloor n \sqrt 7 \rfloor$ is a square. So, are there infinitely many of these?

     38    100 = 2^2 * 5^2
     46    121 = 11^2
     64    169 = 13^2
     97    256 = 2^8
    167    441 = 3^2 * 7^2
    183    484 = 2^2 * 11^2
    200    529 = 23^2
    218    576 = 2^6 * 3^2
    318    841 = 29^2
    437   1156 = 2^2 * 17^2
    490   1296 = 2^4 * 3^4
    546   1444 = 2^2 * 19^2
    575   1521 = 3^2 * 13^2
    605   1600 = 2^6 * 5^2
    667   1764 = 2^2 * 3^2 * 7^2
    699   1849 = 43^2
    732   1936 = 2^4 * 11^2
    800   2116 = 2^2 * 23^2
    835   2209 = 47^2
    871   2304 = 2^8 * 3^2
    945   2500 = 2^2 * 5^4
   1062   2809 = 53^2
   1316   3481 = 59^2
   1361   3600 = 2^4 * 3^2 * 5^2
   1453   3844 = 2^2 * 31^2
   1597   4225 = 5^2 * 13^2
   1697   4489 = 67^2
   1748   4624 = 2^4 * 17^2
   2070   5476 = 2^2 * 37^2
   2241   5929 = 7^2 * 11^2
   2359   6241 = 79^2
   2419   6400 = 2^8 * 5^2
   2480   6561 = 3^8
   2604   6889 = 83^2
   2667   7056 = 2^4 * 3^2 * 7^2
   2731   7225 = 5^2 * 17^2
   2861   7569 = 3^2 * 29^2
   2927   7744 = 2^6 * 11^2
   2994   7921 = 89^2
   3130   8281 = 7^2 * 13^2
   3340   8836 = 2^2 * 47^2
   3630   9604 = 2^2 * 7^4
   3780  10000 = 2^4 * 5^4
   4010  10609 = 103^2
   4247  11236 = 2^2 * 53^2
   4657  12321 = 3^2 * 37^2
   5086  13456 = 2^4 * 29^2
   5174  13689 = 3^4 * 13^2
   5263  13924 = 2^2 * 59^2
   5443  14400 = 2^6 * 3^2 * 5^2
   5534  14641 = 11^4
   5626  14884 = 2^2 * 61^2
   5906  15625 = 5^6
   6290  16641 = 3^2 * 43^2
   6586  17424 = 2^4 * 3^2 * 11^2
   6686  17689 = 7^2 * 19^2
   6787  17956 = 2^2 * 67^2
   6991  18496 = 2^6 * 17^2
   7198  19044 = 2^2 * 3^2 * 23^2
   7303  19321 = 139^2
   7729  20449 = 11^2 * 13^2
   7947  21025 = 5^2 * 29^2
   8057  21316 = 2^2 * 73^2
   8279  21904 = 2^4 * 37^2
   8618  22801 = 151^2
   8848  23409 = 3^4 * 17^2
   8964  23716 = 2^2 * 7^2 * 11^2
   9676  25600 = 2^10 * 5^2