Why do elliptic curves have bad reduction at some point if they are defined over Q, but not necessarily over arbitrary number fields?
Here are two answers: (a) If you try to write down an elliptic curve $y^2 = x^3 + a x + b$ with everywhere good reduction, you need to choose $a$ and $b$ such that $4a^3 + 27 b^2 = $ a unit. We can certainly solve this equation over some (lots!) of number fields, say if we set the unit equal to $1$ or $1$, or a unit in some fixed base number field. But we can't solve it in ${\mathbb Q}$. [Edit: As Bjorn intimates in his comment below, one has to be a little more careful than I am being here to be sure of good reduction mod primes above 2; the details are left to the interested reader (or, I imagine, can be found in Silverman in the section where he discusses the proof that there are no good reduction elliptic curves over $\mathbb Q$).] (b) There are many nontrivial everywhere unramified extensions of number fields (e.g. $\mathbb Q(\sqrt{5}, i)$ over $\mathbb Q(\sqrt{5})$), but there are no everywhere unramified extensions of the particular number field $\mathbb Q$. The situation with elliptic curves is completely analogous. 


There were some discussions on this question. This property is very specific to $\mathbb Z$. To construct elliptic curves with everywhere good reduction over a number field, you can start with any elliptic curve over $\mathbb Q$ with integral $j$invariant. Then it is wellkwnon that $E$ has good reduction everywhere over some finite extension $K$ of $\mathbb Q$ (it is actually enough to take $K$ be the extension generated by the $3$torsion points of $E$). The existence of varieties with good reduction everywhere over number field holds also for abelian varieties and curves of any genus. This can be seen as existence theorem of integral points in some moduli schemes parametrizing abelian schemes and curves over $\mathbb Z$ (look for works of R. Rumely and of L. MoretBailly on Skolem properties). 


It's also worth noting that "good reduction at $\mathfrak{p}$" is a local condition, so an elliptic curve may have everywhere good reduction, despite not having a Weierstrass equation that has good reduction at all primes. This is because over fields of class number greater than 1, there always exist elliptic curves that do not have global minimal Weierstrass equations. The existence, or lack of, a global minimal Weierstrass equation is governed by a certain ideal class (see Proposition VIII.8.2 in my Arithmetic of Elliptic Curves). The fact that if the class number is greater than 1, then there always exist curves with no global minimal Weierstrass equation is in the paper: "Weierstrass equations and the minimal discriminant of an elliptic curve", Mathematika 31 (1984), no. 2, 245–251. There is also a paper by Bekyel that describes the density of curves having (or not having) global minimal Weierstrass equations: "The density of elliptic curves having a global minimal Weierstrass equation", J. Number Theory 109 (2004), no. 1, 41–58. The moral is that you can produce curves with everywhere good reduction by writing down a specific Weierstrass equation, but to determine whether a given curve has everywhere good reduction is done via local calculations, and the associated elliptic scheme having everywhere good reduction may need to be patched together using more than one Weierstrass equation. 


Here is another answer, certainly overkill, but possibly interesting overkill. By the elliptic modularity theorem, the result follows from the fact that the modular curve $X(1)$ has genus $0$: or equivalently, that the upper halfplane modulo $\operatorname{SL}_2(\mathbb{Z})$ is just the affine line $\mathbb{A}^1_{/\mathbb{C}}$. The fact that this curve has to have genus $0$ follows from Fontaine's theorem that there are no curves of positive genus with everywhere good reduction over $\mathbb{Q}$. On the other hand, given a totally real field $F$ of narrow class number one with $[F:\mathbb{Q}] = 2n+1$, we may form the Shimura curve $X_F(1)$ corresponding to a quaternion algebra $B/F$ which is unramified at every finite place of $F$ is ramified at $2n$ out of the $2n+1$ real places; such a quaternion algebra exists because $2n$ is even. This curve does have everywhere good reduction over $F$. Moreover, there is a finite, known list of examples where the genus of this curve is equal to zero (as $F$ ranges over all totally real fields). So  assuming that there are infinitely many such $F$ of strict class number one, which is certainly a widely believed conjecture  one has infinitely many modular curves over totally real fields $F$ with everywhere good reduction. To get elliptic curves out of them one needs rank one factors splitting off from the Jacobian. Again, it seems very plausible that this will happen infinitely many times as we vary $F$ even over all totally real cubic fields of class number one. 


For those with an algorithmic bent, you should look at the paper by Cremona and Lingham "Finding All Elliptic Curves with Good Reduction Outside a Given Set of Primes" In Experimental Math. Volume 16, Issue 3 (2007), 303312. http://www.warwick.ac.uk/~masgaj/papers/egros.pdf 


I would like to stress something that, I think, has not been point out very explicitly in any of the nice answers above. The question that you are asking, if I understand it correctly, is a local question. So let's start with, say, a $p$adic field $K$, and with the equation $f(x,y)=0$ of (the affine piece of) an elliptic curve $E$ with coefficients in the ring of integers $R$ of $K$. To define the reduction of $E$ modulo the maximal ideal $\mathfrak{m}$ of $R$, one cannot follow the naive approach of simply considering the reduction mod $\mathfrak{m}$ of the equation $f(x,y)=0$. Instead one should look at what Tate called a minimal Weierstrass equation $f_{\rm min}(x,y)=0$ for $E$ (LNM 476 pag. 39), and define the reduction $\bar E$ of $E$ as the object obtained by considering the reduction mod $\mathfrak{m}$ of $f_{\rm min}(x,y)=0$. The point now is that the minimal equation is NOT stable by taking field extensions: if $L/K$ is a finite extension then $f_{\rm min}(x,y)$ need not still be a minimal equation for the base change of $E$ to $L$ (but it is if $L/K$ is unramified). Therefore it might very well happens that while $f_{\rm K, min}(x,y)=0$ has a singular reduction mod $\mathfrak{m}$ the corresponding thing does not hold for $f_{\rm L, min}(x,y)=0$ (the meaning of the subscripts is obvious I hope). I tried to come up with an explicit example illustrating this phenomena. I failed to find one in a reasonable amount of time. But I am sure they can be found (easily?) in the literature. 

