There are well enough good answers here, but no one's suggested this method yet, so I'll add it.
Method $1$ - (Secant Approximation)
We can take two simple perfect squares that straddle $2017$. We're just getting a rough estimate, so we'd prefer numbers to be easy to work with than near $2017$. For example, $1600\leq 2017\leq 3600$. The secant line passing $(1600,40)$ and $(3600,60)$ should approximate $\sqrt{x}$ in the interval $[1600,3600]$. This gives us: $\sqrt{x}=24+\frac{x}{100}+\varepsilon_1(x)$. *
So, $\sqrt{2017}\approx44$. Finally, we can check the squares of $43,44,45,\ldots$ and find our answer.
Method $2$ - (Mean of $2$nd Degree Taylor Polynomials)
We can make method $1$ slightly more accurate, though it's not really vital to do so. The $2$nd degree Taylor expansion of $\sqrt{x}$ at $x=a$ is $T_a(x)=\sqrt{a}+\frac{x-a}{80}+\frac{(x-a)^2}{8a\sqrt{a}}$. Then the mean of $T_{1600}$ and $T_{3600}(x)$ should be a good estimate for $\sqrt{x}$ near the centre of the interval $[1600,3600]$. Then $\sqrt{x}=60+\frac{x-2600}{80}-\frac{1}{2\cdot8}\left(\frac{(x-1600)^2}{40^3}+\frac{(x-3600)^2}{60^3}\right)+\varepsilon_2(x)$.** Hence, $\sqrt{2017}\approx45$ and we can check the squares of nearby integers, as in method $1$.
These methods would also work for any $2000<x<3000$ and could easily be adapted for other values of $x$. They also give a convenient way of finding an initial guess, for methods such as Newton-Raphson (detailed in @mathreadler's answer).
Accuracy of Methods
* The error term reaches its maximum at $\varepsilon_1(2500)= 1$. So, in the worst case scenario, we'd need to check the $3$ numbers $(y-1),y,(y+1)$, where $y$ is the estimate of $\sqrt{x}$ from method $1$ and where $x\in[1600,3600]$.
** For $x\in(1769,3110)$, we have $\varepsilon_2(x)<0.67<\varepsilon_1(x)$ but for $x<1769$ or $x>3110$, we have $\varepsilon_2(x)>\varepsilon_1(x)$. In other words, method $1$ is more accurate than method $2$ in the centre of $[1600,3600]$, but the opposite is true near the bounds of the interval. However, since we're interested in the centre of the interval, this is good. The error term, $\varepsilon_2(x)$, reaches a minimum of $0$ around $x=2351$.